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Dr. William Murray’s Differential Equations course will help you solve problems from all over biology, physics, chemistry, and engineering. Dr. Murray demonstrates his extensive teaching experience with easy to understand theory and a wide array of examples and helpful tips. Each lesson also includes several completely worked out practice problems like the ones you will see on homework and tests. Sample topics include linear and separable equations, second order equations, series solutions, systems of equations, and Laplace transforms. Dr. Murray received his Ph.D from UC Berkeley, his BS from Georgetown University, and has been teaching in the university setting for 10+ years.

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I. First-Order Equations
  Linear Equations 1:07:21
   Intro 0:00 
   Lesson Objectives 0:19 
   How to Solve Linear Equations 2:54 
    Calculate the Integrating Factor 2:58 
    Changes the Left Side so We Can Integrate Both Sides 3:27 
    Solving Linear Equations 5:32 
   Further Notes 6:10 
    If P(x) is Negative 6:26 
    Leave Off the Constant 9:38 
    The C Is Important When Integrating Both Sides of the Equation 9:55 
   Example 1 10:29 
   Example 2 22:56 
   Example 3 36:12 
   Example 4 39:24 
   Example 5 44:10 
   Example 6 56:42 
  Separable Equations 35:11
   Intro 0:00 
   Lesson Objectives 0:19 
    Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them 1:33 
    Important to Add C When You Do the Integration 2:27 
   Example 1 4:28 
   Example 2 10:45 
   Example 3 14:43 
   Example 4 19:21 
   Example 5 27:23 
  Slope & Direction Fields 1:11:36
   Intro 0:00 
   Lesson Objectives 0:20 
    If You Can Manipulate a Differential Equation Into a Certain Form, You Can Draw a Slope Field Also Known as a Direction Field 0:23 
    How You Do This 0:45 
   Solution Trajectories 2:49 
    Never Cross Each Other 3:44 
    General Solution to the Differential Equation 4:03 
    Use an Initial Condition to Find Which Solution Trajectory You Want 4:59 
   Example 1 6:52 
   Example 2 14:20 
   Example 3 26:36 
   Example 4 34:21 
   Example 5 46:09 
   Example 6 59:51 
  Applications, Modeling, & Word Problems of First-Order Equations 1:05:19
   Intro 0:00 
   Lesson Overview 0:38 
    Mixing 1:00 
    Population 2:49 
    Finance 3:22 
    Set Variables 4:39 
    Write Differential Equation 6:29 
    Solve It 10:54 
    Answer Questions 11:47 
   Example 1 13:29 
   Example 2 24:53 
   Example 3 32:13 
   Example 4 42:46 
   Example 5 55:05 
  Autonomous Equations & Phase Plane Analysis 1:01:20
   Intro 0:00 
   Lesson Overview 0:18 
    Autonomous Differential Equations Have the Form y' = f(x) 0:21 
    Phase Plane Analysis 0:48 
    y' < 0 2:56 
    y' > 0 3:04 
    If we Perturb the Equilibrium Solutions 5:51 
    Equilibrium Solutions 7:44 
    Solutions Will Return to Stable Equilibria 8:06 
    Solutions Will Tend Away From Unstable Equilibria 9:32 
    Semistable Equilibria 10:59 
   Example 1 11:43 
   Example 2 15:50 
   Example 3 28:27 
   Example 4 31:35 
   Example 5 43:03 
   Example 6 49:01 
II. Second-Order Equations
  Distinct Roots of Second Order Equations 28:44
   Intro 0:00 
   Lesson Overview 0:36 
    Linear Means 0:50 
    Second-Order 1:15 
    Homogeneous 1:30 
    Constant Coefficient 1:55 
    Solve the Characteristic Equation 2:33 
    Roots r1 and r2 3:43 
    To Find c1 and c2, Use Initial Conditions 4:50 
   Example 1 5:46 
   Example 2 8:20 
   Example 3 16:20 
   Example 4 18:26 
   Example 5 23:52 
  Complex Roots of Second Order Equations 31:49
   Intro 0:00 
   Lesson Overview 0:15 
    Sometimes The Characteristic Equation Has Complex Roots 1:12 
   Example 1 3:21 
   Example 2 7:42 
   Example 3 15:25 
   Example 4 18:59 
   Example 5 27:52 
  Repeated Roots & Reduction of Order 43:02
   Intro 0:00 
   Lesson Overview 0:23 
    If the Characteristic Equation Has a Double Root 1:46 
    Reduction of Order 3:10 
   Example 1 7:23 
   Example 2 9:20 
   Example 3 14:12 
   Example 4 31:49 
   Example 5 33:21 
  Undetermined Coefficients of Inhomogeneous Equations 50:01
   Intro 0:00 
   Lesson Overview 0:11 
    Inhomogeneous Equation Means the Right Hand Side is Not 0 Anymore 0:21 
    First Solve the Inhomogeneous Equation 1:04 
    Find a Particular Solution to the Inhomogeneous Equation Using Undetermined Coefficients 2:03 
    g(t) vs. Guess for ypar 2:42 
    If Any Term of Your Guess for ypar Looks Like Any Term of yhom 5:07 
   Example 1 7:54 
   Example 2 15:25 
   Example 3 23:45 
   Example 4 33:35 
   Example 5 42:57 
  Inhomogeneous Equations: Variation of Parameters 49:22
   Intro 0:00 
   Lesson Overview 0:31 
    Inhomogeneous vs. Homogeneous 0:47 
    First Solve the Inhomogeneous Equation 1:17 
    Notice There is No Coefficient in Front of y'' 1:27 
    Find a Particular Solution to the Inhomogeneous Equation Using Variation of Parameters 2:32 
    How to Solve 4:33 
    Hint on Solving the System 5:23 
   Example 1 7:27 
   Example 2 17:46 
   Example 3 23:14 
   Example 4 31:49 
   Example 5 36:00 
III. Series Solutions
  Review of Power Series 57:38
   Intro 0:00 
   Lesson Overview 0:36 
    Taylor Series Expansion 0:37 
    Maclaurin Series 2:36 
    Common Maclaurin Series to Remember From Calculus 3:35 
    Radius of Convergence 7:58 
    Ratio Test 12:05 
   Example 1 15:18 
   Example 2 20:02 
   Example 3 27:32 
   Example 4 39:33 
   Example 5 45:42 
  Series Solutions Near an Ordinary Point 1:20:28
   Intro 0:00 
   Lesson Overview 0:49 
    Guess a Power Series Solution and Calculate Its Derivatives, Example 1 1:03 
    Guess a Power Series Solution and Calculate Its Derivatives, Example 2 3:14 
    Combine the Series 5:00 
    Match Exponents on x By Shifting Indices 5:11 
    Match Starting Indices By Pulling Out Initial Terms 5:51 
    Find a Recurrence Relation on the Coefficients 7:09 
   Example 1 7:46 
   Example 2 19:10 
   Example 3 29:57 
   Example 4 41:46 
   Example 5 57:23 
   Example 6 69:12 
  Euler Equations 24:42
   Intro 0:00 
   Lesson Overview 0:11 
    Euler Equation 0:15 
    Real, Distinct Roots 2:22 
    Real, Repeated Roots 2:37 
    Complex Roots 2:49 
   Example 1 3:51 
   Example 2 6:20 
   Example 3 8:27 
   Example 4 13:04 
   Example 5 15:31 
   Example 6 18:31 
  Series Solutions 1:20:26
   Intro 0:00 
   Lesson Overview 0:13 
    Singular Point 1:17 
   Definition: Pole of Order n 1:58 
    Pole Of Order n 2:04 
    Regular Singular Point 3:25 
   Solving Around Regular Singular Points 7:08 
    Indical Equation 7:30 
    If the Difference Between the Roots is An Integer 8:06 
    If the Difference Between the Roots is Not An Integer 8:29 
   Example 1 8:47 
   Example 2 14:57 
   Example 3 25:40 
   Example 4 47:23 
   Example 5 69:01 
IV. Laplace Transform
  Laplace Transforms 41:52
   Intro 0:00 
   Lesson Overview 0:09 
    Laplace Transform of a Function f(t) 0:18 
    Laplace Transform is Linear 1:04 
   Example 1 1:43 
   Example 2 18:30 
   Example 3 22:06 
   Example 4 28:27 
   Example 5 33:54 
  Inverse Laplace Transforms 47:05
   Intro 0:00 
   Lesson Overview 0:09 
    Laplace Transform L{f} 0:13 
    Run Partial Fractions 0:24 
   Common Laplace Transforms 1:20 
   Example 1 3:24 
   Example 2 9:55 
   Example 3 14:49 
   Example 4 22:03 
   Example 5 33:51 
  Laplace Transform Initial Value Problems 45:15
   Intro 0:00 
   Lesson Overview 0:12 
    Start With Initial Value Problem 0:14 
    Take the Laplace Transform of Both Sides of the Differential Equation 0:37 
    Plug in the Identities 1:20 
    Take the Inverse Laplace Transform to Find y 2:40 
    Example 1 4:15 
    Example 2 11:30 
    Example 3 17:59 
    Example 4 24:51 
    Example 5 36:05 
V. Review of Linear Algebra
  Review of Linear Algebra 57:30
   Intro 0:00 
   Lesson Overview 0:41 
    Matrix 0:54 
    Determinants 4:45 
   3x3 Determinants 5:08 
   Eigenvalues and Eigenvectors 7:01 
    Eigenvector 7:48 
    Eigenvalue 7:54 
   Lesson Overview 8:17 
    Characteristic Polynomial 8:47 
    Find Corresponding Eigenvector 9:03 
   Example 1 10:19 
   Example 2 16:49 
   Example 3 20:52 
   Example 4 25:34 
   Example 5 35:05 
VI. Systems of Equations
  Distinct Real Eigenvalues 59:26
   Intro 0:00 
   Lesson Overview 1:11 
   How to Solve Systems 2:48 
    Find the Eigenvalues and Their Corresponding Eigenvectors 2:50 
    General Solution 4:30 
    Use Initial Conditions to Find c1 and c2 4:57 
   Graphing the Solutions 5:20 
    Solution Trajectories Tend Towards 0 or ∞ Depending on Whether r1 or r2 are Positive or Negative 6:35 
    Solution Trajectories Tend Towards the Axis Spanned by the Eigenvector Corresponding to the Larger Eigenvalue 7:27 
   Example 1 9:05 
   Example 2 21:06 
   Example 3 26:38 
   Example 4 36:40 
   Example 5 43:26 
   Example 6 51:33 
  Complex Eigenvalues 1:03:54
   Intro 0:00 
   Lesson Overview 0:47 
    Recall That to Solve the System of Linear Differential Equations, We find the Eigenvalues and Eigenvectors 0:52 
    If the Eigenvalues are Complex, Then They Will Occur in Conjugate Pairs 1:13 
   Expanding Complex Solutions 2:55 
    Euler's Formula 2:56 
    Multiply This Into the Eigenvector, and Separate Into Real and Imaginary Parts 1:18 
   Graphing Solutions From Complex Eigenvalues 5:34 
   Example 1 9:03 
   Example 2 20:48 
   Example 3 28:34 
   Example 4 41:28 
   Example 5 51:21 
  Repeated Eigenvalues 45:17
   Intro 0:00 
   Lesson Overview 0:44 
    If the Characteristic Equation Has a Repeated Root, Then We First Find the Corresponding Eigenvector 1:14 
    Find the Generalized Eigenvector 1:25 
   Solutions from Repeated Eigenvalues 2:22 
    Form the Two Principal Solutions and the Two General Solution 2:23 
    Use Initial Conditions to Solve for c1 and c2 3:41 
   Graphing the Solutions 3:53 
   Example 1 8:10 
   Example 2 16:24 
   Example 3 23:25 
   Example 4 31:04 
   Example 5 38:17 
VII. Inhomogeneous Systems
  Undetermined Coefficients for Inhomogeneous Systems 43:37
   Intro 0:00 
   Lesson Overview 0:35 
    First Solve the Corresponding Homogeneous System x'=Ax 0:37 
   Solving the Inhomogeneous System 2:32 
    Look for a Single Particular Solution xpar to the Inhomogeneous System 2:36 
    Plug the Guess Into the System and Solve for the Coefficients 3:27 
    Add the Homogeneous Solution and the Particular Solution to Get the General Solution 3:52 
   Example 1 4:49 
   Example 2 9:30 
   Example 3 15:54 
   Example 4 20:39 
   Example 5 29:43 
   Example 6 37:41 
  Variation of Parameters for Inhomogeneous Systems 1:08:12
   Intro 0:00 
   Lesson Overview 0:37 
    Find Two Solutions to the Homogeneous System 2:04 
    Look for a Single Particular Solution xpar to the inhomogeneous system as follows 2:59 
   Solutions by Variation of Parameters 3:35 
   General Solution and Matrix Inversion 6:35 
    General Solution 6:41 
    Hint for Finding Ψ-1 6:58 
   Example 1 8:13 
   Example 2 16:23 
   Example 3 32:23 
   Example 4 37:34 
   Example 5 49:00 
VIII. Numerical Techniques
  Euler's Method 45:30
   Intro 0:00 
   Lesson Overview 0:32 
    Euler's Method is a Way to Find Numerical Approximations for Initial Value Problems That We Cannot Solve Analytically 0:34 
    Based on Drawing Lines Along Slopes in a Direction Field 1:18 
   Formulas for Euler's Method 1:57 
   Example 1 4:47 
   Example 2 14:45 
   Example 3 24:03 
   Example 4 33:01 
   Example 5 37:55 
  Runge-Kutta & The Improved Euler Method 41:04
   Intro 0:00 
   Lesson Overview 0:43 
    Runge-Kutta is Know as the Improved Euler Method 0:46 
    More Sophisticated Than Euler's Method 1:09 
    It is the Fundamental Algorithm Used in Most Professional Software to Solve Differential Equations 1:16 
    Order 2 Runge-Kutta Algorithm 1:45 
   Runge-Kutta Order 2 Algorithm 2:09 
   Example 1 4:57 
   Example 2 10:57 
   Example 3 19:45 
   Example 4 24:35 
   Example 5 31:39 
IX. Partial Differential Equations
  Review of Partial Derivatives 38:22
   Intro 0:00 
   Lesson Overview 1:04 
    Partial Derivative of u with respect to x 1:37 
    Geometrically, ux Represents the Slope As You Walk in the x-direction on the Surface 2:47 
   Computing Partial Derivatives 3:46 
    Algebraically, to Find ux You Treat The Other Variable t as a Constant and Take the Derivative with Respect to x 3:49 
    Second Partial Derivatives 4:16 
    Clairaut's Theorem Says that the Two 'Mixed Partials' Are Always Equal 5:21 
   Example 1 5:34 
   Example 2 7:40 
   Example 3 11:17 
   Example 4 14:23 
   Example 5 31:55 
  The Heat Equation 44:40
   Intro 0:00 
   Lesson Overview 0:28 
    Partial Differential Equation 0:33 
    Most Common Ones 1:17 
    Boundary Value Problem 1:41 
   Common Partial Differential Equations 3:41 
    Heat Equation 4:04 
    Wave Equation 5:44 
    Laplace's Equation 7:50 
   Example 1 8:35 
   Example 2 14:21 
   Example 3 21:04 
   Example 4 25:54 
   Example 5 35:12 
  Separation of Variables 57:44
   Intro 0:00 
   Lesson Overview 0:26 
    Separation of Variables is a Technique for Solving Some Partial Differential Equations 0:29 
   Separation of Variables 2:35 
    Try to Separate the Variables 2:38 
    If You Can, Then Both Sides Must Be Constant 2:52 
    Reorganize These Intro Two Ordinary Differential Equations 3:05 
   Example 1 4:41 
   Example 2 11:06 
   Example 3 18:30 
   Example 4 25:49 
   Example 5 32:53 
  Fourier Series 1:24:33
   Intro 0:00 
   Lesson Overview 0:38 
    Fourier Series 0:42 
    Find the Fourier Coefficients by the Formulas 2:05 
   Notes on Fourier Series 3:34 
    Formula Simplifies 3:35 
    Function Must be Periodic 4:23 
   Even and Odd Functions 5:37 
    Definition 5:45 
    Examples 6:03 
   Even and Odd Functions and Fourier Series 9:47 
    If f is Even 9:52 
    If f is Odd 11:29 
   Extending Functions 12:46 
    If We Want a Cosine Series 14:13 
    If We Wants a Sine Series 15:20 
   Example 1 17:39 
   Example 2 43:23 
   Example 3 51:14 
   Example 4 61:52 
   Example 5 71:53 
  Solution of the Heat Equation 47:41
   Intro 0:00 
   Lesson Overview 0:22 
   Solving the Heat Equation 1:03 
   Procedure for the Heat Equation 3:29 
    Extend So That its Fourier Series Will Have Only Sines 3:57 
    Find the Fourier Series for f(x) 4:19 
   Example 1 5:21 
   Example 2 8:08 
   Example 3 17:42 
   Example 4 25:13 
   Example 5 28:53 
   Example 6 42:22 

Hi and welcome to the differential equations lectures here on www.educator.com.0000

My name is Will Murray and I’m very glad that you are going to be joining me, today we are going to start right away with linear equations, that is that first topic of differential equations.0004

There is a special algorithm on that and we are going to learn to solve them, let us jump right in.0015

We are going to learn how to solve linear differential equations, the way you recognize a linear differential equation is that you can put it in this form Y′ of x + p(x) × y(x)=x3.0020

You might be using different variables like you might have a (t) instead of (x), that is ok the methods are exactly the same.0034

I want to make a couple of notes about the form here, the word linear does not necessarily mean what you think it would mean.0042

You can still have lots of functions in here that are not linear, linear means that we think of y and Y′ as the variables, not x and y.0053

But you think of y and Y′ as the variables and you think of the p(x) and the x3 even though they are functions of (x), you think of those as being the coefficients.0061

They might be something like cos(x) or ex or anything like that, but the point is you think of those as being coefficients.0072

If you look at this equation, the linear form is (Y′) + p × y =q, if you think of y and Y′ as the variables and you think of (p) and cubed being coefficients then what you have there is a linear function in terms of Y′ and (y).0079

That is what we mean when we talk about a linear differential equation.0103

If you have something in front of the Y′(x), if you have a coefficient right here, the solution method that we are going to learn in the next slide does not work until you get rid of the coefficient.0109

What you have to do is divide both sides by that coefficient to get rid of it and give your self a nice, clean Y′(x) with no extra coefficient.0123

Make sure to divide any coefficient here in front of the Y′ before we start the algorithm that we are going to learn on the next slide.0133

Divide away any coefficient here first, that is a very common mistake that some of my own students make when I teach them differential equation.0150

I will give them an equation where I have a coefficient here and they will go through and use the algorithm that we are about to learn but it would not work because they forget to divide that coefficient away.0162

Let us start to learn that algorithm, how you solve linear equations is you calculate something called the integrating factor which means you look at the coefficient of the y(x) which is the (p).0172

You look at the p(x) and you integrate it, you take its integral and you raise (e) to that power.0186

So it is e to the integral of p(x) and we call that the integrating factor i(x) and we multiply that by both sides.0194

A lot times this makes the equation get a lot more complicated, let me show you why we want to do that.0202

What we have here on the left hand side, we have i(x) × Y′ and then here we have y × a very large mess but what this mess turns out to be is the derivative of i(x).0210

When you take the derivative of something, the derivative of (e)u, d by d (u) of (e)u is just (e)u × the derivative of (u).0231

Let me just write d of eu is e u, d(u).0247

What we have is that is the eu right there, this is the derivative of that (u) because the derivative of the integral of p(x) is equal to the original p(x).0256

That sounds a little complicated, I think it will make a lot of sense after we jump in to some more examples.0273

Do not worry about it too much right now, the important thing is to notice that we have i(x) × Y′ and (i)prime(x) × Y.0278

If I write it this way (i)Y′ + (i)′ (y), that exactly what we would have gotten if we have taken the derivative of (i) × (y) using the product rule.0290

We are really exploiting the product rule from calculus 1 to solve this kind of differential equation.0300

What that means is you can integrate both sides, when you integrate both sides you can get i(y) on the left and whatever you get on the right.0308

The point is we are going to take it from there and we are going to solve for (y).0318

I think this will make more sense after we do some examples, so if it is still a little fuzzy right now, do not worry because we will work through some examples together and you will definitely get the hang of it.0322

As I said from the previous side, after you integrate both sides, you are going to get i(x) × y(x) × y on the left.0333

Remember I said you will get i(y) on the left and you will get whatever you get by doing this integration on the right.0341

You can then solve for y(x) by diving both sides by this i(x).0350

That is the plan for these differential equations, after we do some examples I think it will be a lot more clear.0357

There are couple of notes that I want to make before we jump in into some examples, let me just remind you of the original form we are solving here.0367

Y′ + p(x), (y) is equal to q(x) that is the form that we are solving.0377

If p(x) is negative, meaning if you have a negative sign right there, you want to make sure to include that when you are finding i(x), that is a very common mistake that my students often make when I’m teaching differential equation.0386

Remember that the integrating factor i(x) is e to the integral of p(x) d(x).0398

If there is a negative sign here, then that is a part of the p(x) and you have to include that in here, but a lot of people do not notice that negative sign and they forget to include it.0408

Trust me that will really ruin all the calculations after that, make sure you include the negative sign.0418

Another related mistake that is very common is resolving this e to the integral.0425

A lot of times when you take the integral you get the natural log or something, my students sometimes see e to the –natural log or something, I wrote cucumber here, e to the -natural log of cucumber.0433

My students will say “oh I know that e and natural log cancel each other out, so e to the- natural log of cucumber is just –cucumber”.0444

That is wrong, let me and let me show you why it is wrong.0453

If you have e to –natural log of cucumber, then that is really the same, remember the rules of natural logs as e to the natural log of cucumber to the -1 power.0456

The rules of natural logs say you take the numbers on the outside and they turn into the exponents.0476

That is e to the natural log of cucumber-1 which is 1/cucumber, which is just 1/cucumber.0481

That is a very common mistake that I see students make all the time, they want to cancel (e) and natural log but they do it so quickly that they do not notice that negative sign is in there.0496

They just try to pull that negative sign outside, it does not work, this is how it works out as 1/cucumber, you got to be really careful about stuff like that.0508

Another note that I want to make is that when you are doing this integral to find the integrating factor, (e) to the integral of p(x), you can leave off the constant there.0518

It does not really matter if you add an arbitrary constant or not, you might as well leave it off, you do not have to worry about the constant there.0526

However, later on in the problem when you are integrating both sides of the equation, the constant is very important, I can not stress this enough, I need to highlight that part right there.0535

The constant is very important when you are integrating both sides of the equation and it is also very important that you do it exactly at the moment when you do the integration.0546

When you do the integration that is when you have to add the constant.0556

The reason you have to be so careful about that is after you do the integration, you usually have to do a couple more steps of algebra to solve for (y).0560

You have to keep track of that constant as you go through those algebraic steps, sometimes that constant gets tangled up in the equation.0570

This is a very big difference from when you first learned how to do integrals back on calculus 2 or even in calculus 1.0578

You could do the whole problem and then just hack on an arbitrary constant at the end.0588

You could even do 5 problems and at the end you just go back and put +c in every single one of them.0591

That does not work in differential equations, you have to add the constant in the middle of the problem when you do the integration step and then keep track of it after that.0598

If you just go back at the end and tack on a constant at the end, you will get the wrong answer.0608

I think you will see in some examples how that plays out but let us just remember to add the constant exactly when we do the integration.0615

Let us jump in with an example now, we are going to find the general solution to the following differential equation Y′ + xy=x3.0630

Let us remember the strategy here, our generic form for a linear differential equation is Y′ + p(xy)=q(x).0639

That is what we have got here, we got Y′ + a function × (y)=another function.0652

The strategy here is to find the integrating factor i(x)=e to the integral of p(x) (dx), that is always our strategy for linear differential equations.0658

In this case, the p(x) is that (x) right there, so that is the (p), and we have to do (e) to the integral (x)(dx) and that is just e to the x2/2, the integral of (x) is x2/2.0674

When you are doing this integration to find i(x), you do not have to worry about the constant here, so no (+c) necessary at this integration.0690

I’m misspelling necessary here, no (+c) necessary at this step, later on there is going to be another step where we are going to do integration.0704

At that point it is going to be absolutely crucial to include the (+c).0715

What we do with this integrating factor is that we multiply it by both sides of the equation, so I’m going to multiply e to the x2/2 × Y′ + I’m going to write this in the middle.0720

(xe) to the x 2 /2 × Y= x3 e to the x 2/2.0735

The whole point of this linear equations and this integrating factor strategy is that this left hand side is supposed to be the derivative of y × i(x) using the product rule, let us look at that.0746

I’m saying that this is the derivative of y × e to the x2/2, that was our i(x), it is supposed to be the derivative of that using the product rule.0761

It is always good to check at this step that it worked, if we took the derivative of (y)e to the x2/2, we get Y′(e to the x2/2).0772

(y) × the derivative of e to the x2/2, which is e to the x2/2 × (x).0783

That right there is i(x), I′(x) and that right there is i(x), we do get something that looks like the aftermath of the product rule.0790

That confirms that we did our work correctly and this is still equal to x3 e to the x2/2.0805

We are going to have to solve this and the way we are going to that is by integrating both sides, if I integrate the left hand side, I have the derivative so I’m going to go (y) e to the x2/2.0816

That is by integrating both sides is equal to the integral of x3 e to the x2/2 (dx).0831

Now I have to solve that integral on the right, I think I have to take another side to do that because it is not the easiest integral in the world.0843

Let me just recap what happened on this side here, we started out with this linear differential equation in this form Y′ + p(xy)=x3.0850

Our strategy for linear differential equation is to form this integrating factor where you do e to the integral of p(x).0861

P(x) is just (x) and its integral is x2/2, we do not have to add a constant for that integral.0869

We multiply that by both sides, so we multiply that by all three terms here.0876

The point is that makes the left hand side into i(x) × Y′ + I′ (x) × (y), we get a product rule going on and that is the derivative of (y) × i(x).0882

If we integrate both sides, on the left hand side we just get y × i(x), on the right hand side we have an integral that we still need to solve.0895

That is what I’m going to carry over on to the next slide and we will keep going there.0903

We still have a couple of steps to go on example 1 here, we had on the previous side y × e to the x2/2 was equal to the integral of x3 e to the x2(dx).0910

We have to solve that integral and what I’m going to do is make a little substitution here, I’m going to do (u)=x2/2 and then my (du) would be equal to 2x/2, just x(dx).0930

I’m going to write this integral as, I think I’m going to take that x3 and write it as x2 and will give me another (x) to put over here with the (dx).0949

This is eu and now x2 is 2 × (u), so I’m going to pull out 2 out here and change that x2 to a (u), eu and x(dx) is (du).0961

I still need to do integration by parts to solve (ue)u, let me set up integration by parts to solve (ue)u.0976

I have a little short hand technique for integration by parts, it is called tabular integration.0986

I thought this on some earlier educator videos for the calculus 2 or calculus BC series.0991

If you go back and look at those calculus BC lectures here on www.educator.com, you will see a section for integration by parts.0998

That is where I thought this section on this trick for tabular integration to do integration by parts quickly.1007

What I’m going to do is take derivates of (u), derivative of (u) is 1 and derivative of 1 is 0, then I’m going to take integrals of eu, that is just eu and eu again.1014

Then I write little diagonal lines here and put alternating signs positive, negative down the diagonal lines.1028

Then I just multiply down the diagonal lines, I still have a 2 on the outside and I have (u)eu - eu, that is still equal to (y)e to the x2/2.1035

I want to really emphasize something right here, this is the step where I did the integration and this is where I have to put a (+c), that is really crucial here, I will put a (+C) right now.1051

Let me really emphasize that, must add (c), we can not just add (c) at the end, so we must add (+c) when we do the integration, not earlier and not later.1064

The reason we have to do it then is because the next step is to solve for (y) and that is going to involve some algebra that will going to change the equation around a little bit.1092

And the (c) is going to be tangled up in with that algebra and we need to let that happen, we can not just tack it on at the very end when we are all done.1102

What we have here is I need to get things back in terms of (x), 2 × (u) is x2/2, e to the 2/2 – e to the x2/2 + (c).1110

I would like to divide both sides by e to the x2/2 because I want to solve for (y) on the left.1133

If I divide both sides by e to the x2/2, I’m just going to get 2 × in the inside here x2/2 – 1 + (c).1140

Now dividing by e to the x2/2 is the same as multiplying by e to the –(x)2/2.1153

I get (y) is equal to, if I distribute that 2 I get x2 - 2 + a constant × e to the –x2/2.1161

That as far as I can take this one, I’m going to stop there, that is called a general solution right there.1173

The general solution always has an arbitrary constant somewhere in it and the way you would find the value of that arbitrary constant will be the U’s initial conditions if you were given an initial condition.1179

In this case, we are not given an initial condition, we can not find the arbitrary constant, so U’s initial condition if given.1191

In this case, we are not given one to find the exact value of (c) and if you do not have an initial condition then you just leave it in terms of the general solution like what I did here.1209

What we really found here is the general solution, that as far as we can take it on this one.1225

The next example we will have an initial condition and you will see how we are going to plug that in the end to find the value of (c) but in the mean time let us go back over what happened with this example.1236

This is the integral that I got from the previous side, this is all work that we did on the previous side, you can check back on the previous side if you do not remember where this all came from.1247

We solved it to the point of getting (ye) to the x2/2 is equal to this integral and this is a slightly messy integral so I made a little substitution.1259

(u)=x2/2, of course I have to make my d(u) as well and then because I have an x3 here, I factor that in to x2 and an (x).1266

The point of that was I want to segregate out an (x) to be my d(u) here, so that x(dx) became the d(u).1278

That x2 is 2 × (u), that gave me a 2 and a (u) here, we have eu.1283

Just to do (ue)u I have to do a little integration by parts problem.1294

I have this tabular integration method that I covered back in my lectures in the calculus BC section here on www.educator.com.1299

You do derivatives of (u), so this are derivatives and integrals of eu, we do integrals.1308

You cross multiply them along with these little alternating signs plus, minus, plus and so we get (ue)u - eu, substitute back in (u) as x2/2.1316

We are trying to solve for (y) that means I have to divide both sides by e to the x2/2.1331

That is what I did here, I cancelled off my e to the x2/2 here and here.1338

Now with the (c), I have to multiply by e to the –x2/2, that is the same as dividing by e to the x2/2.1342

I just distributed this 2, cancelled of the ½ and we got x2 - 2 + (c)e to the –x2/2.1350

At this point, if you had an initial condition, you would use it now to figure out the value of that arbitrary constant.1359

If you do not have an initial condition which is what happened with this problem, you will just stop and you say that is your general solution.1366

In example 2, we have to find the solution to the following initial value problem cos(XY)′ + sin(xy)=cos(x5) sin (x), and the initial condition is y(0)=2.1377

Let me remind you the format for linear differential equation, it has to be in this form (Y)′ + p(xy) is equal to q(x), it got be in that form.1390

In particular, it is not allowed to have anything, any coefficient in front of the Y′, here we do have a coefficient in front of the Y′, we have cos(x).1403

What we are going to do is divide both sides by that cos(x) to get it into the required linear form, we get Y′ and we are going to divide by this.1414

We got to divide both sides here so I get Y′ + sin(x)/cos(xy), now if I have cos5 on the right and I divide by 1 cos, it will go down to cos4(x) sin (x).1427

I’m not going to worry about the initial condition because paradoxically you use the initial condition at the very end of the problem, we would not worry about that for quite a long time.1445

Let us remember our general strategy for linear differential equations which is that you use this integrating factor i(x) is equal to e to the integral of p(x) d(x).1454

In this case, my p(x) is the coefficient of (y) so that is p(x) right there.1470

By the way, you could rewrite this as tan(x) if you wanted, I do not think that is going to be useful for what is coming up.1475

I do not think there is any particular benefit for writing this as tan(x) but it would not be wrong, that is the p(x) right there.1483

What we get here is the integrating factor i(x)= e to the integral of p(x), so e to the integral of sin(x)/cos(x)d(x).1494

In order to integrate that, I think I want to do a little substitution, of course you could remember what the integral of tan(x) is.1508

Certainly that would be great if you did remember it, but I’m not going to assume that you remember it because we are going to solve that just with a little substitution.1515

We are going to use (u)=cos(x) and then d(u) would be –sin(x)d(x), this is (e) to the integral of, there is a negative sign there so I have to put –d(u).1523

d(u) is sin(x) d(x) with a negative, cos(x) is (u), e to the negative integral of d(u)/(u) and that is e to the –natural log of (u), which is e-natural log of u is cos(x).1541

Now a couple of points to mention here, first you do not have to add the arbitrary constant when you are finding the integrating factor.1559

You can omit the (+c) at this point, it is ok, later on it will be very important to add on that arbitrary constant when we do another step of integration later, but here you do not have to worry about it.1567

Next thing to be very careful about is, let me write this in red to show that it is wrong.1580

A lot of my students would look at this and say this is equal to e to the- natural log of cos(x), the e in the natural log is cancel so this is just –cos(x).1587

That is wrong, I want to make it clear that you do not want to do that, instead what you want to do is you write that as e to the natural log of cos(x)-1.1596

Remember that negative becomes an exponent.1614

That is just cos(x)-1 which is 1/cos(x), that is the actual integrating factor, that is also the same sec(x). 1617

That is the integrating factor not –cos(x), that would have really stirred you wrong.1637

Let me go back to the original equation here, we have Y′ + I think I’m going to write it as tan(x), tan(xy)=cos4 x sin(x).1642

Remember the point of the integrating factor is you are going to multiply both sides by that integrating factor, so multiply both sides by sec(x).1658

We will get sec(x) Y′ + sec(x) tan (xy) is equal to, now remember sec(x) is 1/cos(x), we are really dividing by a cos(x).1667

We will go down to cos3x sin(x) and the point of that is that the left hand side is exactly what you would have gotten if we have taken the derivative of sec(x) × (y) using the product rule.1684

The product rule says derivative of 1 × the other, the derivative of other × 1 and that would have given us sec(x) × Y′ and then the derivative of sec is sec(tan) × (Y).1708

That is exactly what we have gotten with the product rule, have we taken the derivative of sec(x) × (y).1723

The product rule suppose to work for every linear differential equation that is the whole point of this, this is still equal to cos3x sin(x).1731

We are going to integrate both sides here, and on the left that means we are going to get sec(x) just × y without the derivative because we integrated the derivative, we just get back to the original function.1744

On the right we still have the integral of cos3x sin(x) d(x) and we have to wrestle with that integral.1758

I think I’m going to go ahead and work that out on the next slide, let me remind you what happened on this slide.1769

We had this differential equation here but it is not quite in perfect linear form and the reason it is not in perfect linear form is because of this coefficient in front of the Y′.1777

We are not supposed to have anything in front of the Y′ except for a 1, we have to divide a way by that cos(x) so that is why I divided the cos(x) in front of the Y.1789

I got sin(x)/cos(x) and cos5 went down to cos4 because I divide by cos, we got this new formula in differential equation and that is in linear form.1800

We can use our generic linear strategy which is to use this integrating factor which means you look at the p(x) and you try to integrate it and you raise e to that power.1812

To integrate sin(x)/cos(x), I use a little (u) substitution, (u)=cos, d(u)=-sin, so we got –d(u)/u which gave me –natural log(u).1827

I did not have to add (+c) here, that is going to come later on, the –natural log(u) substitutes back to –natural log of cos(x).1839

It is very tempting at this point and it is something that students often do is to try to cancel the e in the natural log.1849

But it does not work because of that negative in between them, you have got to turn that negative into an exponent first then you can cancel off the (e) in the natural log.1856

You will get 1/cos(x) or sec(x) as your integrating factor.1866

The point of that was you multiply both sides of the equation by that integrating factor that is what I’m doing here.1871

On the left I get sec(x) × Y′ + sec(x) tan(x) × y and the point of that is that it is the derivative of sec(x) × y using the product rule.1877

If you use the product rule and expand this out you get exactly these two complicated terms.1889

On the right hand side when I multiply by sec(x), that is like dividing by cos(x), I got cos3x × sin(x).1896

I’m still stuck with that, I still have to integrate that, that is why we are going to jump over on the next slide and keep going.1904

On the previous side, we got it down to sec(x) × (y) was equal to the integral of cos3x × sin(x), this is all coming from the previous side here (dx).1912

I have to solve this integral on the right, I’m going to use a little substitution now, u=cos(x) and my d(u), we always have to find the d(u) when you do a substitution.1928

D(u) then would have to be –sin(x) d(x) and I have the integral of u3 and that sin(x) should be negative.1938

I will put negative out here, -d(u), that is –u4, using the power rule over 4.1949

I just did my integration, now is when I have to add a constant, it is very important that I add it right now, -u4/4, which is -1/4 × changing (u) back to cos(x), this is cos4x + the constant.1959

This is still equal to sec(x) × (y), now I want to solve for (y), I want to divide sides by sec(x), that is the same as multiplying by cos(x).1983

I will multiply both sides by cos(x) to cancel off my sec and on the left I will just get (y) and on the right I will get -1/4 cos5x + (c) × cos(x).1995

That is why we had to add that (c) is because the algebra that we do afterwards to solve for (y) starts to tangle up the (c) into the equation here.2011

If we did not have an initial condition, we have been done because what we found right here is the general solution.2021

But this time we have an initial condition, y(0)=2 and we are going to use that here, y(0)=2, I’m going to plug in 2 for (y) and 0 for (x) -1/4 cos5 of 0 + (c) × cos(0).2027

Cos(0) is 1, 2 is equal to ¼ + (c) and my (c) if I move that ¼ over the other side, 2 + 4 is 9/4.2046

I’m going to plug that back in to my general solution right there and I get y=-1/4, cos5x + 9/4 cos(x).2058

That is my complete solution to the original initial value problem, it solves the differential equation and it matches the initial condition.2076

Let me remind you of the steps here, it was on the previous side where we got this down to the point of having sec(x) × (y)= the integral of cos3x sin(x).2087

To solve that integral we used a little (u) substitution, (u)=cos(x), du=-sin(x), because of that negative, that is where that negative comes from.2100

Then we get an easy integral of u3, that means u4 + a constant, on the left hand side I have this sec(x) × y.2108

To solve for (y), I’m going to divide by sec(x), that is the same as multiplying by cos(x), I multiplied both sides by cos(x), that is why this 4th became a 5th here and the (c) got multiplied by cos(x).2119

Now I use my initial condition here, plug in 0 for (x) and 2 for (y), so 2 is -1/4 cos50 + (c) cos(0).2134

Cos(0)=1, 2=-1/4 + a constant, move the ¼ to the over other side I get (c)=9/4 then I take that and I plug it back in to my general solution right here, that is where I get y=-1/4 cos5x + 9/4 cos(x).2145

Example 3, we are given a linear differential equation and we are just asked to find the integrating factor.2173

We do not have to solve this one all the way through, what we have to do here is just find the integrating factor.2178

Remember the general form for linear differential equation though is Y′ +p(xy)=q(x).2187

In particular, you are not allowed to have anything in front of the Y′, any coefficient here, we do have a coefficient right here, x+ 1.2197

What I’m going to do is divide that away, I’m going to go Y′ – 1/x + 1y=1/x + 1 × sin(x).2206

The way you find the integrating factor, you do (e) to the integral of p(x)(dx), the p(x) here is whatever in front of the Y and the key thing here is do not forget the negative sign.2223

The negative is part of it, so that is all p(x) right there, this is e to the integral of -1/x + 1(dx), now that is (e) to the -, now the integral of 1/x + 1 is the natural log of x + 1.2237

You do not have to worry about a constant when you are finding the integrating factor.2261

You can not cancel the (e) and the natural log because there is a negative sign in between them so you have to write (e) to the natural log of x + 1-1.2266

Now you can cancel the (e) in the natural log, you will get x + 1-1 and that is just the same as 1/x+1.2278

That is all we are asked to do for this example, is to find that integrating factor and we found it, we are done.2291

Let me go over those steps one more time, the key thing here is I remembered the generic form for a linear differential equations.2299

It got to be Y′ + p(xy)=q(x) and that only works if the coefficient of Y′ is 1.2307

Here, the coefficient of Y′ was x+1, I just divide it away, I just divided both sides by that, that is how I got it into this form Y′ -1/x + 1y = 1/x + 1 × sin(x).2314

Whatever that is in front the Y is your p(x) but you got to include that negative sign.2330

I plugged that into my generic integrating factor formula including the negative sign and that is an easy integral.2337

The integral of 1/x + 1 is just natural log of x + 1, but there is still that negative sign so I can not cancel the (e) and the natural log yet.2343

Instead I make that -1 in to an exponent and now I can cancel the (e) and the natural log, I end up with just 1/x + 1.2351

In example 4, we are given a linear differential equation sin(XY)′ – cos(xy)=x2 and we want to find the integrating factor i(x).2366

Let us remember our generic form for our linear differential equation, Y′ + p(xy)=q(x).2377

In particular, you can not have any coefficient in front of the (Y)′ so we got this sin(x) here, we got to get rid of it by dividing that away.2387

Divide that and get Y′ – cos(x)/sin(x), y=x2/sin(x), and now I can use my form for the integrating factor which is i(x)=e to the integral of p(x)(dx).2397

That p(x) means whatever in front of the y but it includes the negative sign so go ahead and put that in there (e)cos(x)/sin/(x)(dx).2424

To solve that integral, I could think of that as cot(x) but I do not think that is particularly useful here, what I’m going to do is a little substitution.2438

I’m going to use (u)=sin(x) and d(u) is –cos(x)(dx), (du) is just positive cos(x) because the derivative of sin is +cosine, I was looking ahead at my negative sign there but the derivative of sin is just +cos.2449

We have (e)-integral of (du)/u, that negative sign stays there but the integral (du)/(u) is –natural log(u).2481

That is (e)-natural log(u) was sin(x) and we got this negative sign in between the (e) and the natural log, I can not cancel.2499

A lot of students are very tempted to cancel the (e) and the natural log and say “that is –sin(x)”, but it is not.2511

Instead, you use that negative and make it into an exponent that is (e)natural log(sin x) to the -1.2518

Now you can cancel the (e) and the natural log, you got sin(x)-1 and that is equal to 1/sin(x) which is cosec(x).2527

By the way, sin(x) of -1 is not the same as arcsin or inverse sin(x), this is really an exponent so it really means 1/sin(x).2544

Be very careful here but this is not the same as sin -1 (x) or arcsin(x), it does not mean the same thing as either one of those, that is another frequent mistake that students make.2554

We are done here because we are just asked to find i(x) and we found it, let me recap the steps there.2570

We wanted to get the equation into linear form Y′ + p(xy)= q(x), in order to do that, I have to divide away this sin(x).2577

I divided away and I get Y′ – cos/sin(y)= x2 /sin(x) and in the integrating factor you do (e) to the integral p(x)(dx), where the p(x) is whatever is found in front of the Y.2587

You got to include that negative sign there, that is all p(x).2601

To do that integral, I did a little substitution (u)=sin(x), d(u)=cos(x)d(x), we got –(du)/u which is –natural log(u).2607

That is substituting back, that is –natural log(sin x) now I can not cancel the (e) and the natural log yet because I still got the negative sign in between them.2618

I will make that an exponent and now I can cancel the (e) and the natural log and I will just get sin(x)-1, that is not the same as inverse sin or arcsin, it is really sin(x)-1.2627

That is 1/sin(x) which is cosec(x).2644

On example 5, we got to solve the following initial value problem t + 1y′- 3y=t and y(1)=2.2652

Let us go ahead and get that into linear form because it is not quite in linear form yet.2664

Remember linear form is Y′ + p(t), y=q(t) I have got to divide away that t + 1 to get it in that form.2670

I’m going to go Y′ – 3/t + 1y=d/t + 1, I’m not going to worry about that initial condition for quite a long time yet, you do not have to worry about that.2682

My integrating factor, that is my strategy for all linear differential equations is (e)integral of p(x) or p(t) in this is case, which is (e)integral of -3/t +1 p(t).2696

Remember I got to include the negative sign when I’m finding the integrating factor.2716

This is (e)-3 d(t)/t + 1, I use a little substitution (u)=t + 1, d(u)=d(t), then I just got 1/d(u)/(u), this is (e) -3 natural log (t +1).2724

This is (e)natural log(t +1) raised to the -3 power, now I can cancel my (e) and my natural log.2751

That is t + 1-3 which is just 1/(t + 1)3.2764

Now the point of that integrating factor is you got to multiply it by both sides of the differential equation.2776

We will multiply that by both sides of this form of the differential equation, I get 1/(t + 1)3 × Y′ – 3/(t + 1)4 × y= t + 14.2785

The point of that is that this left hand side is always equal to the integrating factor × y if you took their derivative using the product rule.2811

It is always a good exercise to check whether that left hand side really is the derivative of i(t) *y2824

Because it is a derivative what we can do is integrate both sides and we will get i(t) × y= integral of t/t +14 d(t).2833

My i(t) remember was 1/t + 13 × y = integral of t/t + 14 d(t).2851

I have to solve that integral and that looks a little bit unpleasant, there is a nice little trick you can do here, which is (u) substitution.2867

(u)=t +1 and then d(u) would just be equal to d(t), I see I’m going to have a (t) in the numerator that I have to deal with.2873

That would be t=(u-1), if I solve my substitution for (u), this is equal to the integral of (u-1) × d(u) all over (u)4.2884

I can separate that out in to (u)/(u)4 which is (u)-3 - (u)-4(du).2901

Now I can integrate using the power rule, the integral of (u)n(du), learned this back in first semester calculus is (u)n +1/n + 1.2915

If (n) is -3, (n) + 1 is -2, this is (u)-2/-2, now (n) is -4, (u)-3/-3.2929

I just did the integration, now is the step when I have to add the arbitrary constant, not sooner not later, it got to be right now.2949

(+c) here, if I turn that back in to t + 1, I’m going to pull this -2 out and make it -1/2, t + 1-2 + 1/3, t + 1-3 + c.2957

This is all equal to 1/t + 13 × (y), and now I want to solve for (y) so I’m going to multiply both sides by t + 13 on both sides here.2984

On the left I get (y) = on the right – ½, (-2 + 3) is just 1 so, (t + 1) + 1/3.3004

T + 1-3 × t + 13 is just 1 + (c) × (t + 13).3020

I’m going to keep going with this in this next slide so we can solve for the constant but let me remind you what we did in this side.3029

I had to get it in this generic linear equation form which means I have to get rid of this coefficient of (t +1).3037

I divide both sides with (t + 1) and this was my new form of the differential equation.3044

I looked at whatever was in front of the Y and I call that p(t), I have to integrate that.3050

I got to integrate -3/(t + 1) that is easy with the (u) substitution, (u)=(t + 1) so I get -3 natural log(u), natural log(t + 1).3056

Before I can cancel my (e) and my natural log, I have to take that -3 and turn it into an exponent then I can cancel the (e) and the natural log.3069

And I will get (t + 1)-3 or 1/(t + 13).3078

I took that 1/(t + 13) and multiply that by both sides 1/(t + 13) and that is how I get this line here 1/(t + 13) Y′, 3/t + 14y and t/t + 14.3081

The point of that integrating factor, this is always true for all linear differential equations, the left hand side turns in to the derivative of a product.3106

It is the derivative of the integrating factor × y, the right hand side you will never going to know what it turns in to but we integrate to undo that derivative.3116

We got the integral of the right hand side t/(t + ¼), the i(t) is 1/(t +13).3126

On the right hand side I have got to integrate t/(t + 14), my clever idea there is to use (u)=t + 1 and d(u)=d(t), plug that in for (u) here, plug (u -1) in for (t).3136

That now separates out into two nice powers of (u), well fairly nice powers of (u), they are negative.3150

I just remembered the power rule from calculus 1 here, I plug that in with n=-3, n=-3, and n=-2 on the right and that was the step where I did my integration.3156

I have to add a (+c) right at that step then I changed my u’s back to (t + 1) and I just brought this down 1/(t + 13) × y.3174

In order to solve that I multiplied both sides by (t + 13) and that is how on the right I got -1/2(t + 1)1 and 1/3 (t + 1)0 and (c) × (t + 13).3187

We are not done with this yet because we have not yet incorporated the initial condition, what I’m going to do is take this equation over on the next slide.3202

We are going to use it to solve the initial condition and then we are going to be done.3210

Let me remind you of the equation that we had on the previous slide.3216

Our solution so far is (y)=-1/2 × (t +1) + 1/3 + (c) × (t + 13).3222

Now I’m going to use my initial condition y(1)=2, I plug in 2 for (y) and 1 for (t), so 2=-1/2 × if (t) is equal to 1 then I will get a 2 there + 1/3 + (c) × 23.3238

2=-1 + 1/3 + it looks like it is going to be 8(c) there and if I move that -1 over, I get 3 =1/3 + 8(c).3261

If I move that 1/3 over, 3 is 9/3, so 3 – the 3rd is 8/3 is equal to 8(c) and I see I will get (c)=1/3.3279

I take that (c)=1/3 and I plug it back into my general solution and I get (y)=-1/2 *(t +1) + 1/3 + 1/3 × (t + 13).3292

I can do some algebra here, I could maybe expand out (t + 13), maybe expand out -1/2 (t + 1).3313

I do not think it will be dramatically better so I’m going to leave that in that form, I have figure out what my arbitrary constant is, I have solved this as far as it is really useful to go.3322

Let me remind you of the steps here, this came from the previous slide, this is our solution from the previous slide.3338

But it was not a complete solution yet because we have not incorporated the initial condition.3344

We use the initial condition to find the value of the arbitrary constant (c), I’m plugging in from here I get y=2, from here I get t=1.3349

I plugged in y=2 there and t=1 in here and here, that is why my (t + 1) turned into two’s, that simplified to -1, 1/3, 23 is 8, I just did a little algebra to simply down and figure out that (c) was 1/3.3361

I take that and plug it back into the general solution, that is where I got this 1/3, not this 1/3, but that is where I got this 1/3 and the general solution is the arbitrary constant.3381

My whole thing turns into just the general solution with the arbitrary constant given the value of 1/3.3393

On example 6, we are going to find the general solution to the following differential equation (XY)′ + 3y=cos(x) with (x) being positive here.3403

That is going to be a linear differential equation but we need to do some manipulations to get it in to the right form.3419

Remember that form for linear differential equation is Y′ + p(xy)=q(x).3424

In particular, there is not any coefficient in front of the Y′, which is what we have here, I’m going to divide that away to start, Y′ + 3/(xy) = cos(x)/x.3434

That is why we are given the condition (x) is positive, it allows me to divide by (x), I do not have to worry about dividing by 0 there, that is really all we use that condition for.3450

Now I’m going to use my integrating factor strategy for linear equations, my i(x)= (e) to the integral of p(x) d(x) and my p(x) is whatever is in front of the y.3462

That is the p(x) right there, in this case the p(x) is positive 3/x, I get (e) to the integral (3/x d(x)).3482

The integral of 3/x is just 3 natural log(x), I do not need to add (+c) when I’m just finding the integrating factor.3497

A lot of people would like to say that is 3x because it is (e) to the 3 log (x), not true.3506

You can not cancel (e) to the natural log until you get rid of that 3 in between them, so I’m going to make that into an exponent on the (x).3512

(e) to the natural log of (x)3 and this is just x3.3519

What you do with that integrating factor is you multiply both sides by whatever you got.3526

I’m going to multiply both sides by x3 and I’m going to get x3 Y′ + (3/x × x2) is 3x2, y= x2 × cos(x).3533

The point of using that integrating factor is that the left hand side is supposed to look like a derivative.3549

In this case, it looks like the derivative of x3 × y, the reason for that is because if you expand out the derivative of x3 × y using the product rule,3557

You get x3 × the derivative of (y) + (y) × the derivative of (x)3, so it really does check there. 3568

It is always a good strategy to check when you are at this step in any linear differential equation because what it does is it just make sure that you did all the previous work right.3575

It is really easy to make mistakes through this process, it is good to check and make sure that the derivative of x3y really is equal to what you had on the left hand side.3587

In this case is does work, this is still equal to x2 cos (x) and in order to undo that derivative we want to take the integral of both sides.3598

We get x3y the integral of the derivative is just the original thing is equal to x2cos(x) or the integral of x2 cos (x) d(x).3608

In order to integrate that, I have to use parts and I’m going to use my tabular integration trick that I taught you back in the calculus BC lectures here on www.educator.com.3622

You can look this up on this lectures if you like, remember the trick is to take derivatives of the left hand side until you get to 0.3634

The derivative of x2 is 2x, derivative of 2x is 2, derivative of 2 is 0, then you take integrals of the right hand side.3650

The integral of cos(x)=sin(x), the integral of sin(x)=-cos(x), integral of –cos(x)= -sin(x).3655

Then I cross multiply down this diagonals and I put alternating signs plus, minus, plus, and I’m ready to write the answer to my integral.3665

It means you multiply down this diagonal lines, you will get x2sin(x) + 2x cos(x) – 2 sin(x).3677

I have adjusted my integrals, now is the time to add my arbitrary constant, the key is you add it this time and the reason is because we still have not finished solving for (y).3692

As we solve for (y), we are going to do some algebraic manipulations and the (c) is going to be tangled up in there algebraic manipulations.3703

You have to let it get tangled up in those algebraic manipulations or other wise it is the wrong answer3709

In this case, to solve for (y) I’m going to divide both sides by x3 so I get y=x2 sin(x) + 2.3718

It looks like I made a small mistake there I forgot to right down that (x), so it is actually 2x cos(x) – 2 sin(x) + (c).3729

That is all divided by x3 because I was trying to get rid of that x3 on the left hand side.3750

That is as far as I can take it for this one, the reason is because I do not have an initial condition, all I have is the differential equation.3756

If I had an initial condition I would use that to find the constant, I would use that at this step.3768

Let me write that down, you would use the initial condition if it is given to you to find the value of (c).3773

You can narrow that (c) down to a particular number if you have that initial condition but in this case we did not have an initial condition so we just stop here with our general solution.3793

General solution means that it has an arbitrary constant in the answer.3805

Let me recap the steps here, I was remembering my generic form for a linear differential equation, it does not have anything in front of the Y′.3813

It is not a 0 there, let me not suggest that supposed to be a 0 but it just does not have anything there, it is like its multiplied by1.3823

In this case we have an (x) in front of the Y′, the first thing I have to do right away is divide everything by (x).3831

I divide this (x) away, at least away from the Y′ term, that means it shows up in the denominator over here 3/x and cos(x)/x.3837

Now I can use my generic solution method for linear differential equations which is to use the integrating factor, which is (e) to the integral of p(x).3847

That p(x) is the 3/(x), it is whatever in front of the Y if there is a negative sign, you got to include that as well.3857

In this case, there is no negative sign so we have (e) to the integral of 3/x, the integral of 3/x is 3 natural log(x).3864

Before I can cancel the natural log I have to move that 3 over to be an exponent.3875

I get x3 or (e) to the natural log x3 but the (e) in the natural log now they do cancel, I get x3.3880

I take that and I multiply that by both sides of the equation and that is how I get x3 Y′ + 3x2y=x2cos(x).3887

The point of that is that the left hand side is the derivative using the product rule of x3y.3900

We are really using the product rule there and because it is a derivative of something I can integrate both sides.3908

Integral of x3 Y′ is just equal to x3y and on the right I have the integral of x 2 cos(x) which is a little nasty.3915

I’m going to need integration by parts by that but I have my little tabular method that I have learned back in the calculus 2 or calculus BC lectures here on www.educator.com.3925

Here is my tabular integration method, I’m taking derivatives of x2, so I got down the 2x, 2, and 0.3936

I’m taking integrals of cos(x), the integral of cos is sin, the integral of that is -cos, the integral of that is –sin.3945

Then I multiply down these diagonal lines attaching alternating signs as I do it, I’m multiplying alternating signs on.3952

I get x2 sin(x) I see I have 2 negatives here so that gives me + 2x cos(x) – 2 sin(x).3961

It is very important that you have the constant when you do the integration which is right now, I put a constant on there.3972

Because I have x3y, I want to solve for (y) so I just divided that x3 in to the denominator there.3979

If I have an initial condition I would use that to find the constant but since I do not, I just stop here and I present my general solution to my differential equation.3987

That is the end of this lecture on linear differential equations, we got a whole lot more interesting material to cover in differential equations.3999

This is just the beginning of the topic, I hope you will stick around and sign up to see the other lectures on differential equations.4005

We are going to talk about separable differential equations in the next lecture.4013

We are getting in to all kinds of other topics, systems of differential equations, partial differential equations, second order differential equations.4017

We are going to graph slope fields and direction fields, there is all kinds of exciting material for us to explore together and I really hope you will be a part of it.4025

In the mean time, my name is Will Murray and you are watching the differential equations lectures on www.educator.com, thanks for joining us.4034

Hi welcome back to www.educator.com, I’m Will Murray doing the differential equations lectures.0000

Today we are going to talk about inhomogeneous equations and the method of variation of parameters.0005

There are 2 ways to solve inhomogeneous equations in the last lecture we learned the method of undetermined coefficients and we worked through that very carefully.0011

Today we are going to learn a totally different method which is called variation of parameters, let us see how that works out.0022

To solve the linear second order inhomogeneous, that is really the key word here, inhomogeneous constant coefficient differential equation, Y″ + bY′ + cy=g(t).0032

The key part here is that inhomogeneous, remember homogeneous means you have a 0 on the right hand side, inhomogeneous means that you have a function here that is not 0, a g(t).0046

We are going to solve the homogeneous equation first by the methods of earlier lectures, remember we had several lectures on how to solve homogeneous equations.0058

If you do not remember how to do that, maybe go back and check the earlier lectures in the differential equation series and what you will get is a homogeneous solution c1 y1 + c2 y2.0070

Before we go on I want to emphasize one point here, you will notice here that there is no coefficient in front of the Y″.0083

That is going to be a key point to make some of the equations later on work out.0092

If you are trying to solve one of these equations and you do have a coefficient in front of the Y″, what you want to do is divide that away any coefficient here.0097

Let us see some examples of that and that really does matter and it is something that can really easily lead you into mistakes when you are solving these things, we will be careful with that.0116

First you solve the homogeneous equation, it means you set the right hand side equal to 0, that is what homogeneous means.0127

Using what we learned about the characteristic equation and complex roots from repeated roots, all that business took us several lectures to unfold to get a homogeneous solution c1 y1 + c2 y2.0136

Let me show you where you go from there, you want to find a particular solution to the inhomogeneous using what is called variation of parameters.0151

What that means is where we had c1 before and c2, we are going to replace those by generic functions u1(t) and u2(t).0161

We do not know what the u1 and u2 are but we are going to replace those by functions and that is going to form our particular solution.0174

Here is how you figure out what u1 and u2 are, you solve this system y1 u1′ + y2 u2′=g.0185

Let me stress here what is known and what is unknown, the known is the y1 and y2.0204

You know that because you got those from your homogeneous solution which you already solved.0215

You also know (g), because that came from the g(t), that is how get this g.0224

What you do not know and what you are trying to solve for are the u1 and u2, remember the u stands for unknown here.0232

Since you do not know what u1 and u2 are, you do not know what their derivatives are either and you want to think of the 2 variables here as being u1′ and u2′.0243

We have 2 equations and unknown here, it is like an algebraic system that you solved in high school algebra but it is actually trickier than that,0254

Because all of these things are functions instead of just constants and numbers.0264

Let me walk you through how you would solve these things, you solve the system for u1 and u2′.0270

That just tells you the derivatives of u1 and u2, then you have to integrate u1′ and u2′ to get u1 and u2.0278

Once you figure out those, you plug those in to your original guess for the particular solution and you make those the coefficients of y1 and y2.0289

Remember those will be functions of (t), it is not c1 y1 + c2 y2, they are not constants anymore, they are functions of (t).0298

I want to expand a little bit on this business of solving the system for u1′ and u2′ because that can be a little unpleasant, it is not as easy as solving systems in high school algebra.0309

I will be gave a short hand way of solving that, you got find what is called the Wronskian of the 2 homogeneous solutions.0321

Wronski was a Polish mathematician so the Wronskian is named after him.0331

It means that you set up this little matrix of y1 and y2 and then on the second line you put their derivatives y1′ and y2′.0336

Then you take the determinant of this matrix which means you cross multiply y1 × y2′ and y2 × y1′ and you subtract those.0347

You get this determinant y1 y2′ – y2 y1′.0360

The way you can find u1′ and u2′ is you plug in that determinant that you just figured out in the denominator and the numerator you have y2 and that (g).0368

Remember that was from the right hand side of the original differential equation Y″ + bY′ +cy=g(t) that is where the (g) comes from.0383

And the y1 and y2 come from your homogeneous solutions.0396

That is kind of a formulaic way you can find u1′ and u2′ but you still have to integrate them to get u1 and u2.0402

And then you take those u1 and u2 and plug in to your guess for the particular solution u1 y1 + u2 y2.0423

There are several steps here we will go through some examples so you will get a chance to get more comfortable with this.0440

Let us go ahead and jump in to the examples.0446

We are starting with the homogeneous differential equation (t2 Y″ – t) × (t + 2Y′) + (t + 2y)=0.0448

We are given the general solution there, the homogeneous solution is c1 t + c2 tet and then we are asked to find the general solution to the inhomogeneous differential equation.0460

It is the same stuff on the left hand side but inhomogeneous means the right hand side is not 0.0477

We replaced it with 2tq, let us go ahead and see how that one works out.0483

I wanted to give myself a new clean slate here.0498

Remember a key part of solving these things using variation of parameters is that you can not have a coefficient on the Y″.0501

Here we do have this coefficient so right away I’m just going to divide the whole equation by t2.0510

I get Y″ -, now t/t2 is 1/t, I will just write t + 2/t Y′ + t + 2/t2y = t3/t2 is just t.0518

What I did there was I divided away that t2 so I will have a nice clean Y″ with no coefficient there and that means my g(t) is 2t, that is the g(t).0541

The homogeneous solution is where you get your y1 and y2, so that is y1 and that is y2.0560

I’m going to follow the solution hint that I gave you in the lesson overview where we find the Wronskian of y1 and y2.0569

That means you write the original equations on the first line, y1 is t and y2 is tet.0579

On the second line you take their derivatives, the derivative of t is 1, derivative of tee we got to use the product rule on that.0591

It is t × the derivative of et, so that is tet + et × the derivative of t, that is just et × 1.0600

That is our Wronskian and we need to take the determinant of that, which is t2 et + tet- the back diagonal tet.0621

Those 2 terms cancel out and we get t2 et, that is our Wronskian determinant.0646

Remember our formula for u1′ is –g × y2/ the Wronskian, so that is negative.0655

My g was 2t, 2t × y2 is tet, I divide that by the Wronskian determinant which is t2et.0669

It looks like I have a t2 cancelling and an et cancelling, I will just get -2 there and that was my u1′.0688

I integrate that to get u1, integrate that would respect to t will get 2t.0698

You do not need to add constants when you are doing these integrals, ultimately our particular solution is going to be u1 y1 + u2 y2.0706

If I added a constant to u1, I would end up just getting more constant multiples of y1 in my particular solution which would just replicate terms from my homogeneous solution.0722

There is no point in adding constant to u1 here.0737

U2′, remember our formula back from the lesson overview is (gy) 1/ the Wronskian determinant.0741

Our g is 2t, our y1 comes from here, is t, divided by t2et and I get 2.0754

The t’s are all cancelled but I still have this et in the denominator, that is the same as multiplying by e-t.0771

The integral of u2′, the integral of 2e-tdt which is negative to e-t, not dt anymore because I have done the integral.0783

I do not need to add a constant when I do that integration, my particular solution is u1 y1 + u2 y2 which is -2t, that is u1 × y1 is t + u2 is -2 e-t and y2 is tet. 0801

This actually simplifies down a bit, we get -2t2, the e-t and et cancel each other out (-2t).0838

Let me point something out, this -2t part is actually this replicates a term of the Y homogeneous there.0850

It means that it is really not contributing anything to the particular solution.0874

If you ran it through the differential equation you just get 0 because it is a homogeneous solution.0877

We can omit, we can leave that part off.0886

Which means when we go for the general solution I can leave that part off, it is the homogeneous solution + the particular solution just like before.0896

It is c1 t + c2 tet , and now my particular solution I can leave off the -2t because it is already built in to this part -2t2.0908

We have solved our very first problem using variation of parameters.0931

Let me just recap how that worked out here, first thing we have to do is get rid of this t2.0942

We did not want any coefficient in front of the Y″ so I divided it out and that meant that the term on the right hand side became 2t, so that was the g(t) that I used there.0947

Then I wanted to use my formulas for variations of parameters, first thing I have to do for that was to build this Wronskian matrix.0960

Which means I take the 2 functions from the homogeneous solution t and tet, write down their derivatives and get a 2 by 2 matrix, take its determinant by cross multiplying.0968

That simplify down to t2 et and then I used my 2 formulas for u1′ and u2′ which reduced down to -2 and 2e-t.0982

Integrate then to get u1 and u2, then I plugged it back into our guess for the particular solution which means that you take u1 and u2 and you multiply them by y1 and y2. 0995

That is what I’m doing here, I noticed that one term of the particular solution was actually a copy of this term of the homogeneous solution. 1011

I decided to leave it out because it is already built in to the homogeneous solution, I leave that part out and I added my particular solution back to the homogeneous solution to get my general solution.1021

At this point, if I have initial conditions I would use them to find the values of c1 and c2.1040

In this case I was not given an initial condition, I’m not going to find c1 and c2.1054

I’m just going to stop with my general solution.1060

Let us move on, for the inhomogeneous differential equation Y″ + y= tan t, we want to solve the corresponding homogeneous equation and find the Wronskian of the solutions.1067

The homogenous equation means you take the right hand side and you set it equal to 0, that is Y″ + y is equal to 0.1084

The characteristic equation, this is actually one that sometimes trips students up because they will say r2 + r=0.1095

That is not correct, remember the r term corresponds to the Y′ term not the y term, that is incorrect.1104

Let me show you, the correct one would be r2 + 1 =0, you will get r2=-1, r= i or -i.1115

Remember we have a formula back when we got complex roots to the characteristic equation, you can look that up a few lectures back when we have homogeneous equations.1131

There was a lecture called complex roots, it is said that when you have a + bi then you r solution is c1 eat cos.1142

I think we might have listed cos first, we might have listed sin first, it does not matter you can list either one first, + c2 eat cos(bt).1161

In this case our a is 0 and our b is 1, because we got 0 + 1i and 0-1i, these ea t terms that just gives you e0.1176

Those terms just turn into 1 and our homogeneous solution is c1 sin(1)t, just c1 sin(t) + c2 cos(t).1192

We got a nice homogeneous solution, it is not the solution to the inhomogeneous equation yet but it is a nice homogeneous solution.1214

We are also asked to find the Wronskian of the solutions, remember we set up this little matrix sin(t) cos(t) and then we take the derivative.1228

The derivative of sin is –cos and the derivative of cos is sin.1246

Reverse those, the derivative of sin(t) is +cos(t) and the derivative of cos is –sin(t).1259

To take the determinant we cross multiply there.1269

We get –sin2 and this one is positive but there was a negative in there so we get –sin2. 1274

The back direction you always subtract so –cos2 and we get -1 for the Wronskian.1286

Let me mention here that this is a slightly unusual situation.1300

Normally the Wronskian will be a function of t not just a simple number, this was a nice one and it just simplify down to a number.1307

Normally the Wronskian will be a function of t, it will be et or cos (t) or something like that.1324

We have the Wronskian and we have our homogeneous solution.1331

That is all was have to do for this example but you want to make sure you really understand this example because we are going to use this one to solve the next one as well.1336

Let me recap quickly how this worked out, we wrote down the characteristic equation.1345

This is the correct form of the characteristic equation and that one I was just trying to highlight a common mistake that student make.1350

We found the roots, looked back at our old form for complex roots to find the generic form of the homogeneous solution.1358

This is plugging in a=0, b=1, and we get our homogeneous solution and that was the y1, that was the y2.1368

We set up our little matrix, found the derivatives and cross multiplied, simplified down to -1 as our Wronskian.1380

We are going to use all these to solve the inhomogeneous equation in the next example.1388

In example 3, we are going to keep going from example 2 where we have the inhomogeneous differential equation Y″ + y= tan(t).1395

Let me just remind you that we already solved the homogeneous equation in the previous example.1408

Here is what we figured out in example 2, the homogenous solution is c1 sin(t) + c2 cos(t).1416

That means our y1 is sin(t) and y2 is cos(t).1432

We also figured out the Wronskian of those 2 solutions, the Wronskian matrix turn out the derivative, the determinant turned out to be -11437

We are going to use all that to build our inhomogeneous solution using variation of parameters.1446

Let me show you how that one goes, u1′, here is our generic formula that we had at the very beginning of the lecture its –y2g/the Wronskian.1453

In this case, our y2 we got up above is cos(t) and our g, that is the right hand side of the original differential equation, that is g(t) right there.1469

This is tan(t) and our Wronskian is -1 so the negative is cancel we got cos(t) and tan(t) I’m going to write that as sin(t)/cos(t).1484

The cos is cancel and we just get sin(t), that was just u1′.1502

To get u1, we have to integrate that, the integral of sin(t)dt is –cos(t).1508

Remember, when we do this integration you do not need to add the arbitrary constant, if you did then you would end up just like replicating copies of your homogeneous solution.1518

That is why you do not have to worry about the arbitrary constant.1529

U2′, again using our formula from the beginning of the lecture is y1 g/the Wronskian.1532

Our y1 was sin(t) and our g was tan(t), Wronskian is -1.1543

Now, I put the negative outside, sin(t) and tan(t) is sin(t)/cos(t) and this is –sin2t/cos(t), that is a little uglier.1555

I’m going to write that as –sin2 as the same as 1 – cos 2(t)/cos(t).1577

I will distribute the negative signs so I get cos2t – 1/cos(t).1588

That is the same as cos(t), if I break up the fraction, -1/cos(t) is sec(t).1597

That was all, remember u2′, we still have not actually found u2.1613

To get u2 we take the integral of that, so let me keep going up here.1618

U2 is the integral of cos(t) – sec(t)dt, so we got to integrate that.1622

The integral of cos is sin(t), the integral of sec(t) is really not something you want to figure out on the spot.1635

That is just something that you have to remember from either calculus 1 or calculus 2.1644

If you do not remember that, you might want to check back at the www.educator.com lectures for calculus 2 in particular, we figured out what the integral of sec(t) is.1650

It was the natural log of sec(t) + tan(t).1659

Now we are going to take u1 and u2 and we are going to plug in our generic formula for the particular solution.1676

That is u1 y1 + u2 y2 and u1 is –cos(t) that is from here, y1 from here is sin(t).1685

Now u2 is this long expression sin(t), I will multiply it through by y2 which is cos(t) as I go along, + sin(t) cos(t) – cos(t) natural log of sec(t) + tan(t).1706

That is our particular solution but it works out pretty nicely because you can see we got a –cos(t) sin(t) here and then a +sin(t) cos(t)so those cancel out.1732

It simplifies down to –cos(y) × the natural log of sec(t) + tan(t).1744

That was our particular solution and to get the general solution, remember it is the homogeneous solution + the particular solution.1757

I go back to, my homogeneous solution is c1 sin(t) + c2 cos(t) + -cos(t) × natural log of sec(t) + tan(t).1770

You do not put an arbitrary constant on that particular part.1791

You just have arbitrary constants on the homogeneous part, the particular part was very carefully designed to work exactly right for the inhomogeneous equation.1798

It does not get an arbitrary constant.1808

To summarize what we did here, we started out with our homogeneous solution which we found back in example 2 and the previous examples.1813

You can check back there if you do not remember where that comes from, c1 × sin(t), c2 cos(t), sin(t) is y1 and cos(t) is y2.1823

We had these generic formulas for variation of parameters, u1′ is –y2g/the Wronskian and u2′ is y1g/the Wronskian.1834

For each of those we get our g(t) over here from the right hand side of the equation.1845

We plug in what y1 and y2 are I here and here and work them out.1850

That was u1′ so we had to take its integral to get u1, that was our u1.1856

This was u2′, a little bit of trigonometric cleverness and then we had to take the integral of that to get u2.1866

We plug those back into our generic form for the particular solution u1 y1 + u2 y2.1875

It turned out after we plugged in what u1 y1, u2 y2 where and got a nice cancellation here.1882

That simplified down and gave us a slightly simpler form for the particular solution.1889

We added that back on to the homogeneous solution c1 sin(t) + c2 cos(t).1894

That plus our particular solution gave us the general solution to the differential equation.1901

On our next example, we are going to start with the inhomogeneous differential equation Y″ + 3y′ + 2y=sin(et).1909

We are going to solve the corresponding homogeneous equation and we are going to find the Wronskian of the solutions.1919

The homogeneous equation means that you take the right hand side and you just set it equal to 0, you throw away whatever you see over there.1926

We are going to throw away the sin(et) and we are joust going to get Y″ + 3Y′ + 2y =0.1935

We learned in a few lectures back how to solve this kind of thing, you just set up the characteristic equation r2 + 3r + 2=0.1945

That factors nicely, that is r + 1 × r + 2=0.1957

I get r=-1 or -2, if you did not want to factor it, you could also do this by the quadratic equation.1971

My own personal feeling is that the quadratic equation is slight more susceptible to mistakes than the factoring method, but if you are more comfortable it is certainly ok to do it that way.1979

We get the homogeneous equation is c1 e-t that is from the -1 + c2 e-2t that is from the -2.1991

We got our homogeneous equation and that is a good step toward what we want.2005

What we are asked to do now is to find the Wronskian of the solution.2014

That means that you make a little matrix, e-t and e-2t and then find their derivatives.2019

Well, derivative of e-t is – e-t, derivative of e-2t is -2e-2t.2030

We are going to find the determinant of that matrix and that means we cross multiply here.2039

Positive direction there gives us -2e-t × e-2t, remember you add the exponents so that is e-3t.2047

In the negative direction we cross multiply and we get -e -2t, e-t, so that is e-3t.2062

But then we are subtracting it, so -e-3t, that turns into a plus and we get -2 +1, -e-3t as our Wronskian.2073

That is what we got by solving the inhomogeneous equation and finding the Wronskian of the solutions.2092

You want to hang on to this because the next example is to solve,2098

Sorry that was for solving the homogeneous equation.2102

We are going to use this same equation for the example and we will solve the inhomogeneous equation.2105

Let me recap quickly how we did this, we started with the homogeneous equation which means you just throw away the right hand side, set it equal to 0.2112

Solve that using the characteristic equation, factor it down, get a couple of roots, the roots become the exponents, the coefficients and the exponents, we get our homogeneous equation.2122

That is going to be the y1, that is going to be the y2, and then we set up our little matrix here with y1 and y2 and the derivatives of each one.2134

We multiply the positive forward direction, we multiply the negative backwards direction, add them up and we get our Wronskian there.2147

Hang on to these solutions, we are going to use them in the next example.2157

For example 5, we are going to solve the inhomogeneous differential equation Y″ + 3Y′ + 2y= sin et.2162

Let me remind you that we have solved the homogeneous equation for this one in the previous example. 2173

Let me bring you up to date with what we have learned from example 4 here.2180

If you have not looked at example 4 recently, just go back and read that over so this would not be a complete mystery.2188

Example 4 told us that the homogeneous solution to this equation was c1 et + c2 e-2t.2194

That was our y1 and that was our y2 and we already found the Wronskian of those which turned out to be –e-3t.2213

We are in pretty good shape to get started on the inhomogeneous equation here which means we are going to use variation of parameters.2226

Using our formulas for variation of parameters u1′ is –y2 × g/the Wronskian, so –y2 is e-2t.2234

G is sin(et), remember g is whatever you see on the right hand side there, so sin(et), got that over here from g(t).2253

Our Wronskian determinant is –e-3t, this does simplify a bit.2270

First of course the 2 negative cancel each other, this e-3t, if you bring it up into the numerator it becomes e3t.2279

We have e3t, e-2t, just gives you e1t × sin(et), that simplifies a bit.2292

To get u1 itself, that was just the derivative we found there, we need to integrate e1t × sin(et) dt and we got to do a little substitution there.2303

We are going to do a little s substitution, I can not use u, normally I like to use u for u substitutions but we are already using the u’s for something else here.2317

I’m going to use another variable and I picked s, s = et and ds is going to be derivative of e and t is just eedt.2328

Here I have sin(s) and this ee dt just give me a ds and the integral of sin(s) is –cos(s), which is –cos(et),.2340

That was a little substitution, it might be worth reviewing the calculus 1 lectures for substitution here on www.educator.com if you are a little rusty on how to make substitutions in integration.2363

Let me show you, the next step is to figure u2′.2376

Again, using the formulas from the beginning, we have this is y1g/the Wronskian determinant.2380

Our y1 was e-t, g is still sin(et) and we are dividing that by –e-3t, that is the Wronskian determinant.2389

If I simplify that a bit, I will pull the negative outside and I will get e-3t when it comes up to the numerator turns into e3t.2407

We got e3t and et that gives us e2t × sin(et) and that is just u2′.2419

To get u2 I have to integrate that so u2 is the integral of that, couple of negative outside so e2t × sin(etdt).2431

We can solve this one using the same substitution, I will get my s is et, ds is etdt.2447

This part will give me a sin(s) and now this e2t, I’m going to write that as et × et.2462

The reason I’m doing that is I need one et for my (ds), if I take on et to be (ds).2472

Those were the collectively give me my (ds), I still have one et left so there is still an (s) left over there.2482

I’m integrating s sin(s)/(ds) and that is still a little ugly, I’m going to use integration by parts here.2491

That was something we covered in the calculus 2 lectures here on www.educator.com, you can check back and look up your material on integration by parts if you are a little bit rusty on that.2499

I’m going to the quick version, the integral of s sin(s), let me use tabular integration here.2510

I’m going to write derivatives on the left, t he derivative of s is 1, the derivative of 1 is 0.2520

The integral of sin(s) is –cos(s), and the integral of cos(s) is sin, so the integral of –cos(s) is -sin(s)2525

Then I make little diagonal lines, put a + and – and what I get here, I still have this one negative from the outside.2539

I’m going to read along the diagonal lines there, -s cos(s) + because we have couple of minus there that cancel each other sin(s).2550

If I distribute my negative back in, and by the way remember we do not have to include the (c) when we are integrating to get this u’s.2567

That is one nice thing you do not have to worry about, if I distribute my negative back in I will get +s cos(s) and let us remember that s was et.2575

Et × cos(et) –sin (s) because there was a negative on the outside, - sin (et), that was my u2 there.2587

Now I found u1 and I found u2, I’m ready to just plug them in to the generic form for my particular solution which is u1 y1 + u2 y2.2606

U1 is –cos(et and that gets multiplied by y1 which is e-t + u2 is et cos(et) – sin(et).2628

All of that gets multiplied by y2 which is e-2t, what happens here if I distribute that e-2t here.2663

E-2t × et is e-t × cos(et).2675

I have got e-tcos(et) here, I’m not forgetting that other term, we will get to that in a second.2687

But over here we got -cos(et) × e-t.2694

Those 2 terms cancel each other out, this is pretty common in variation of parameters.2699

You will get 2 terms that cancel each other out in some funny way, it is not always obvious when it happen but it is nice when it does.2704

The only term I have left here is coming from these 2 factors, -(e2t) × sin(et), it is the way it simplifies down there.2713

That is my particular solution, my general solution remember is the homogeneous solution + the particular solution.2730

That is the homogeneous solution we already figured out back in example 4 was e-t + c2 e-2t.2746

I’m going to add on the particular solution that we just worked out that is negative, -e-2t sin(et).2756

That is my general solution since I do not have initial conditions, I do not need to figure out what c1 and c2 are.2772

You do not put a constant on a particular solution, just on a homogeneous solution.2781

That is my general solution and I’m done with that one.2787

Let me recap what we did there, we started out with the homogeneous solution which we found from example 4.2792

If that was a mystery where that came from, go back and watch the video for example 4 and you will see where that comes from.2799

Read off y1 and y2 there and we had also figured out the Wronskian matrix in the previous example.2808

That is still coming from example 4, we have our Wronskian there and then we used our formulas for u1′ and u2′ in terms of y1 and their Wronskian matrix and y2.2818

There is a (g) in there, and we get the (g) from the right hand side of the original differential equation, that is where that (g) comes from, here and here.2832

We simplify those down that gives us u1′ and u2′ and to get u1 and u2 themselves you have to integrate those.2842

That is what we are doing here is we are integrating with a little substitution.2853

First one is not a bad integral to get u1 by itself, remember you do not have to add a constant when you are doing these integrals to find u1 and u2.2857

U2 is a little messier, we had to integrate this and we did a substitution first and then even after substituting we had an integral that we needed to solve by integration by parts.2868

That is what I was doing up here, this little table is doing the integration by parts to integrate s sin(s).2880

If that is fussy for you, you might want to go back and check the lectures for calculus 2 here on www.educator.com.2887

There is a whole lecture on integration by parts where you get some practice with that.2894

After we did that integration by parts we got to substitute back and distribute the negative sign (s is et) and we got our u2.2900

We took u1 and u2 and plug them into our generic form for the particular solution, so there is our u1 and u2.2910

Multiply them by y1 and y2, it was nice because some terms cancel, that term cancel with that term here.2919

It simplify down to a single term here, that was our particular solution.2928

To find the general solution remember you add the particular solution back to your homogeneous solution.2934

There is our homogeneous solution again and there is our particular solution being added on to it.2940

That is how variation of parameters works and that is the end of this lecture in the differential equation series.2949

I’m very happy that you are watching, my name is Will Murray and you are watching www.educator.com.2955

Thanks for joining us, bye.2961

Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and we are going to be doing a review of powers series today.0000

The reason is that later on we are going to be learning how to use power series and Taylor series to solve differential equations.0010

Since it might have been a while since you actually studied power series and Taylor series, we are going to work through some of the basic properties today.0018

In the next lecture you will be ready to go to use power series and Taylor series to solve differential equations. 0026

Let us jump right in, here were going to start out with the review of the definition of Taylor series.0034

You start with a function f(x)and you take a point x0 and we find the Taylor series expansion around that point which means the sum, remember the capital letter σ means the sum.0043

You are adding these terms up from n=0 infinity of a sub n, x times first times X minus X0 to the end where the a's of n, these coefficients are fn(x0)/n factorial.0058

Just a couple reminders here remember that this fn what that means is the nth derivative of f and derivative of the function f. 0077

When n=0 that means the zero derivative which means you're taking the derivative zero times which just means the original function f and when n=0 that is just f0 itself, n factorial means you multiply the numbers 1 x 2 x 3 x 4 up to n, that is what n factorial means.0093

Let me remind you of that 1 x 2 x 3 up to n and by convention we say zero factorial is equal to 1.0116

That is something that throws off a lot of my students at first but it makes all the formulas later on work out nicely so it is worth accepting that convention that zero factorial is equal to 1.0132

Most common type of Taylor series that we study here is when x0 is equal to 0 so the the point here x0 would be equal to 0 and that is called the Maclaurin series.0145

Maclaurin series just means a special case of the Taylor series when the x0 is equal to 0. 0158

In that case, the Maclaurin series of s of a function is the sum from n=0 to infinity of a sub n(xn) and the a of n is still the same.0164

It is the sum from n=0 to infinity of the nth derivative of f applied to x0 would be 0 because we are doing the Maclaurin series divided by n factorial x xn.0181

That is the Maclaurin series and that is probably 90% of the examples you study in calculus and differential equations, maybe even more than that or actually involving Maclaurin series which is the special case of the Taylor series. 0200

There are certain Maclaurin series that it is really worth memorizing, hopefully you memorize these back in your calculus class and I will just give you a little reminder of them now.0216

But if you did not memorize back then, it is worth memorizing them at least while you are using series to solve differential equations.0228

Just very common functions are ex sin x cosine x and 1/1 - x.0236

The Maclaurin series for ex is 1+ x + x2/ 2 factorial + x3/3 factorial.0243

The general pattern there is that this is the sum from n=0 to infinity of xn/ n factorial.0252

Remember that if n=0 then you get x0, you just get 1 and then you get n factorial.0263

0 factorial is just 1, this is the n=0 term gives this 1 over here the n=1 term gives the x and then it is a little more obvious in the terms after that and how they correspond to the general sum.0270

The next one that you really want to remember and recognize is sin x which is x - x3/3 factorial + x5/5 factorial -x7 factorial.0286

It is a little alternating series and the way we can make it alternate is we put a -1 to the n term.0300

That makes it alternate the positive and the negative and the way we get the odd powers there is we do x2 n +1.0310

As n ranges overall number 01234, 2 n + 1 will range over the odd numbers 1357.0324

That is the way of collecting those odd powers and we want the same odd numbers in the denominator.0332

We put 2n +1 factorial in the denominator, that is a close form expression for the Maclaurin series for sin x.0340

For cos x, it is quite similar except you have the even powers 1 - x2/ 2 factorial plus x4/4 factorial - x6/6 factorial.0353

Again we can now write this summation form for that, the sum from n= 0 to infinity again it is alternating so I'm going to put it somewhere -1n in there 0365

And then to get the even powers I'm going to go x2n/(2n factorial).0379

Be a little careful there with the 2n factorial, some people think that (2n factorial) is equal to 2 factorial x n factorial or 2 x n factorial, that is emphatically not true.0387

You have to be very careful with the parentheses there 2n factorial, for example of n=10 and then 2n factorial would be 20 factorial which is not the same as 2×10 factorial, you have to be quite careful with the parentheses there.0404

The last series that we are going to use very frequently when we are studying differential equations is the Geometric Series 1/1-x.0419

That expands into 1+ x + x2 + x3, that is the simplest one to write down a formula for, that is just the sum from n= 0 to infinity of xn.0429

If you plug in the n=0 term you will just get that first one there and that is true for the absolute value of x < 1, in other words x between -1 and 1.0443

That is not valid when x is outside that range, outside interval, that it is valid when x is inside that interval.0459

That brings me to the next topic here which is where did these series converge? What values of x are these series are good for?0471

That is something that we study back in Calculus 2, by the way if all of these is feeling more rusty then you can brush off in a single lesson.0481

You might want to go back and look at the Calculus 2 lectures Here on www.educator.com.0489

We spent several lectures back in Calculus 2 going over Taylor series explicitly so you can get at a more lengthy rundown on all these topics back in Calculus 2.0494

But if you remember those pretty well and you just need to brush off a little rust then keep playing along here.0505

Any power series has a radius of convergence around x0 so what that means, let me draw a number line here 0511

The center of the radius of convergence is always that value x0 and then what you do is you go a distance, the same distance in either direction and that is the radius r.0524

That means the top and the interval you have x0 + r and at the bottom of the interval you have x0 - r.0540

The idea is that when you are inside that radius the series converges, in this green region here the series converges.0551

When you are outside that radius this series diverges, what that means is if you plug in values of x that are outside this radius.0566

If you plug those values into the series then the series diverges and this is called the Radius of Convergence.0575

A shorthand algebraic way of keeping track of this is to write this as x - x0 in absolute value is less than r because that will exactly capture this values between x0 - r and x0 + r.0583

By the way, when I say x nought, nought means 0. So that is just x…saying x nought is just a shorthand way of saying x0.0602

The big question here is whether it converges at the end points, x= x0 - r and x=x0 + r and we do not know that just when we find out the radius.0611

The series are quite unpredictable, sometimes it will converge at one end point but not the other one, Sometimes they will converge on both end points, sometimes it will converge at neither one, we do not really know that until we test each one individually.0629

There are some extreme cases to this radius, If r=0 that means that the series converges only at x0 and at nowhere else.0644

That is the case where r=0 and that is really a pretty worthless series because if you think about it if you plug in x=x0, x = x0 that would mean that x - x0 is 0 so you are just getting a series of all 0, 0 to the end and that is really a worthless series.0657

If you add up a bunch of 0 you will certainly get 0, it certainly converges which you do not really learned very much, r=0 can happen but it is essentially a worthless case.0682

The other extreme of that is r=infinity, r = infinity means that the series converges infinitely far in either direction so it converges for all points on either side of x0, that is a very good case.0693

It means you can plug in any value of x you like and the series will converge, that is the best case if the radius is infinite.0714

The way you want to find the radius of convergence of a taylor series or have any power series, it is usually to use the ratio test now back in Calculus 2 we learn a whole lot of different task.0726

We learned the alternating series test, the integral test, the ratio test, and root tests and all these tests in Calculus 2.0739

When you come right down to Taylor Series and Power Series the one you use almost all the time is the ratio test, that is the only one that I'm really going to mention here. 0747

But if you want some more details on checking when series converge, you might want to go back and check those lectures for Calculus 2.0758

The way you check if a series converges is you form this ratio where you look at the nth term so that means if you have a generic term then you would plug in for that generic term.0767

Then you would plug in n+1 into that generic term, you go back and replace all the n's by n +1 and you divide them together, that is the ratio.0786

That is why it is called a ratio test, you take its absolute value which is nice because it gets rid of any negatives that might be hanging around.0794

You take the limit as n goes to infinity and what you generally get is some expression in terms of x and you set that less than 1.0802

That will tell you for all those values of x wherever values of x you end up solving for.0815

Those are the values for which these series converges and the values of x that make this limit greater than 1 diverge. 0821

The way this looks on a number line is you have x0 and then you will have some region within which that limit will be less than 1.0833

Outside that region I will put this in red here, that limit will be greater than 1 on either end and so it converges when the limit is less than 1.0847

Maybe I will put that in green to emphasise that that is the convergent region, limit here is less than 1 and that is the convergent region.0862

Now the big question here is the end points and the ratio test never tells you what happens at the end points x0 - r and x0 + r.0874

You do not know from the ratio test whether the series converges there and what you have to do is plug both of those values in separately.0886

Each one into the series separately and see whether it converges using some test other than the ratio test.0894

The ratio test is guaranteed to fail at the end points because it will give you 1 and that is the cut off value so instead you use some other ratio, some other test besides the ratio test that we learned back in Calculus 2.0901

Let us see how all this works out, first example here is to identify the following power series as an elementary function.0918

We are given the series 1+ 3x2 9/2x4 9/2 x x6 27/8 x x8 this is a typical kind of thing you will get at the end of the differential equations problem.0926

Where you have a series as your solution and you are trying to recognize that as some function maybe that you have seen before somewhere.0946

Let me show you how to think about this, I see 3x2 is really nothing clever.0953

I can do and that and I see 9x4 and what I noticed is that (3x2)2.0960

I'm going to write it that way and it looks like I'm starting to build up some kind of pattern here, I'm going to write (32)3 for the next term but (3x2)3.0972

That would give me 27 here and all I had was a 9, to balance that out I'm going To divide by 3 in order to balance out the fact that all I had here was 9.0990

I'm going to keep going here, I will put (3x2)4 but 34 is 81 and since I had a 27 here I have to divide by 3 to keep it balanced, that is 3×8 which is 24. 1002

This Is 6, I'm noticing that I can write this in terms of powers of 3x2, as long as I modify the denominators accordingly.1027

I'm going to write this as 1+ 3x2 + (3x2)2, I look at these numbers in the denominator and I see 2, 6, and 4, that is pretty clearly a factorial pattern so 6 is 3 factorial and 24 is 4 factorial.1040

What I recognize here is something that looks like the series that I have memorized for ex , let me write that down just quickly up here.1074

Remember the ex is 1+ x + x2/2 factorial, 2 factorial is just 2 + x3/3 factorial + x4/4 factorial and here 2 factorial is the same as 2.1083

I'm going to write that is 2 factorial, I see that I have exactly the same series here except that I got 3x2 wherever there was a and x in the series for ex.1106

What this gives me is e(3X)2.1123

What I did there was to look at the series that I was given and I noticed that the powers of x are going up by 2, I know that there is something about the power of x2.1139

I also noticed just for the first 2 terms I had 3 and 32 and so I thought maybe I have some pattern and powers of 3 there.1149

That is why I started making patterns of 3x2 all the way through here and then I had to put some extra numbers in the denominator just to make that pattern work out.1159

When I looked at the numbers that I ended up with the denominator, I noticed that I got this factorial pattern 2, 6, 24 is 2 factorial, 3 factorial.1174

Once I got it into that form I remembered my series for ex and I noticed That I have the same series where x has been replaced by 3x21185

That is how I know that this whole thing is equal to 3x2. 1196

Next example we are going to find the Maclaurin series for the function f(x)=natural log of 1-x.1203

There is a very seductive mistake that can be made here that calculus students often make, they use the generic formula for Maclaurin Series, which is to write down the sum for n=0 to infinity of fn of 0/ n factorial x x - x0n.1214

But x0 is 0, it is just xn, I will start writing down derivatives and I will look for a pattern.1239

That can often be the longest and slowest, most difficult way to find a Maclaurin series and the reason is sometimes writing down the 3rd and 4th derivative of a function gets very messy.1247

You are very liable to make mistakes and it is very hard to spot the power so instead of going from that basic formula and just writing down derivatives,1260

I want to introduce another method which is to try to relate this function to one of the ones from which we already have memorized series.1268

What I know is that natural log of x, if I take the derivative of natural log of x, dy/dx of natural log of 1-x.1279

What I get is 1/1-x and in the chain rule x -1 so the point of that is that 1/1-x, I have a memorized prefabricated known Maclaurin series.1293

I'm going to use that Maclaurin series that I already know for 1/1-x and use it to build a series for natural Log of 1-x.1315

Let us see how that works out, formally natural log of 1-x I know I can get that by starting with 1/1-x and taking its integral of one little issue here which is this -1 right here.1323

I will put in negative outside here and I will throw on a dx, I'm going to take the integral of 1/1-x and that is going to give me natural log of 1-x along with a negative sign.1343

The point here is that I know a series for 1/1-x so my series for 1/1-x that was the geometric series 1+ x + x2 + x3 and so on.1358

When we had that back at the beginning of this lecture that was something we figured out long ago in our calculus lectures the series for natural log of 1-x, this is just like a polynomial and it is very easy to integrate.1374

If I integrate this, I have that negative and the integral of 1 is x and the integral of x is x2/2, the integral of x2 is x3/3 and so on.1389

Whenever you do an integral you have to add an arbitrary constant so I will put in a constant on the outside.1406

I'm going to put it on the left here and so this is c - x, I need to distribute that negative sign so it's minus -x2/2, -x3/3 and so on.1414

That is supposed to be equal to natural log one of 1-x and I do not know what that constant is yet so in order to figure out what that constant is I'm going to plug in a value of x.1431

The easiest value of x for me to plug in is x=0 and I'm going to plug that in to both sides, I will plug in x=0 to both sides and on the left the natural log of n=0 , so that is natural log of 1.1442

On the right, I will get c- a bunch of 0 and natural log of 1 i know that that's equal 0, I will get c=0.1472

I know what my c is and i can go back and fill that in and natural log of 1-x= -x - x 2/2 and I dropped out the c to now because i figured out it was equal to 0- x3/3 and so on.1484

I can write that in σ form as the sum from n= it looks like it has to be 1 to infinity because I'm starting at n=1 here.1506

xn/n and there is still a negative sign in there so that is my series for natural log of 1-x my maclaurin series for natural log of 1-x that might be one the you remember doing in calculus 2.1517

Just to recap there what happened was I had to find the maclaurin series and this is my generic formula for maclaurin series for a function.1539

But even though it is tempting to use that formula and start writing down derivatives it is easier to relate the series and function to one of those known functions that we already have a series for. 1550

That is what I try to do over here is I try to remember that the derivative of natural log of 1-x is the same as 1/1-x except for a negative sign.1563

I know the maclaurin series for 1/1-x, I am excluding that fact very heavily so I set up natural log of 1-x is the integral of 1-x and a negative sign convert that 1/1-x.1574

This is using the fact that I already memorized the series for 1/1-x convert that into its series form .1590

That is just a polynomial so it is very easy to integrate I just integrate it term by term and 1 gives me x, the x gives me x2/2, x2 gives me x3/3 and so on.1598

I also have an arbitrary constant so that's my series for natural log of 1-x except that I have not figured out what that constant is.1608

So to find that constant I plug in x= 0 to both side so 0 there and then 0 for all this x's and on the left that gives me natural log of 1which is 0.1617

On the right, that just gives me c, I figure out that c=0.1629

When I plug that back in I get my series natural log of 1-x is -x-x2/2 - x3/3 and so on.1634

I can write that in σ form like this and that is my maclaurin series for natural log of 1-x.1645

In example 3, we are going to find the interval of convergence for that maclaurin series that we just figured out.1653

Let me remind you what that series was, we figured out that natural log of 1-x is equal to the sum from n=1 to infinity of xn/n.1659

Not n factorial just xn/n and it's negative so the way we are going to find the interval of convergence is we are going to look at the ratio test.1677

Remember the ratio test says you look at the n +1 term and divide that by the n term of the n term and the denominator first, because the easier part that is -xn/n. 1695

The n+1 term means you replace all the n's by n +1, so -xn +1/n + 1, that is the limit that we had to simplify.1720

Some good things happen right away which is that because we have absolute values all the negative signs go away. 1736

I do not have to worry about negative sign, I will write that down that it is killed by the absolute value and now we are dividing a fraction by a fraction so I'm going to flip the bottom fraction up.1745

I have n/xn x xn +1, just flipping that bottom fraction up to the top, it is still an absolute value and I'm going to reorganize things a little bit and I'm going to sort all the x's together.1765

xn +1/xn x n/n + 1 and we are supposed to be taking the limit of that and I'm going to separate these two terms xn +1 /xn. 1784

That just simplifies down to x and the limit as n goes to infinity of n/n +1, you can divide top and bottom by the highest power.1804

That is n here so this limit gives me 1 in the numerator and then n/n is 1+1/n and the 1/n goes to 0 so that whole thing goes to 1.1820

This whole thing gives me just the absolute value of x after I take the limit.1840

The idea with the ratio test is that limit is supposed to be less than one for the series to converge, what that tells me is that this is where the series converges.1845

Absolute value of x less than 1 that means x is between -1 and 1 and it diverges outside of that region, do not diverge if x is less than -1, x bigger than 1.1869

The only thing that ratio test does not tell us and it never tells us is whether or not the series converges at the end points.1888

We always have to check the endpoint separately and you can not use the ratio test to check the endpoints ever.1895

We are going to plug x= -1 and x=1 into that series separately and we can not use the ratio test anymore. 1906

We plug in x=1 into that series right here then we will get the sum from n= 1 to infinity of xn will just be 1/n, that is the negative of 1+ 1/2 + 1/3 + 1/4 and so on.1915

This is not something we discussed yet in these lectures on differential equations but it is something we talked a lot in the lectures on calculus two and I'm hoping that it something you remember.1941

If not you might want to go back and check at the lectures on calculus two. but this is the harmonic series that we studied in calculus 2.1950

It is an example of a p series if that rings a bell for you and what we learned in calculus 2 is that, that diverges.1965

The harmonic series diverges, that means that x=1 is not at the interval of convergence because it makes the series diverge.1975

Let us check the other endpoint that means we are going to plug in x = -1 in here into the series, -1 or n=1 to the infinity x -1n.1986

For the harmonic series, when n=1 here we get -11 x -1, we get +1 that -1n essentially gives us the same series except it is going to be alternating, that is going to be 1-1/2-1/3-1/4.2004

This is something we have not studied here in differential equations yet but it is something we studied back in calculus 2.2022

We figured out that this is not alternating harmonic series which does converge by what we call at that time the alternating series test.2032

Go back and look at the lecture in calculus 2 on alternating series and you will see this is actually one of the examples we studied back then.2050

We figured out that x= -1 does make it converge, x =1 makes it diverge.2065

Let me summarize this.2073

x does converge 1x= -1 the series does converge and when x=1 it does not converge that is why I'm putting less than equal to, for x = -1 and less than for x =1, that is my interval of convergence and sometimes you write this in interval notation just different notation for the same thing.2086

We are going to write the interval from -1 to 1and we are going to include the end point at -1, we are going to put square brackets there and we are going to exclude the end point at 1, we will put a round brackets there.2110

Such as different notation for the exact same thing square brackets on one side to show that endpoint is included and round brackets on the other side to show that end point is excluded.2123

By the way, we could have possibly predicted this, I will say that this makes sense if you think about the original function that we studied here which is natural log 1-x.2137

Let us see what happens if you plug in natural log of 1-x, if you plug in the x=1.2157

If you plug in x=1, that is natural log of 1-1 so it is natural log of 0 and if you remember the graph of natural log, it looks like this.2168

At 0, it kinds of blows up to negative infinity, it is undefined.2180

It approaches negative infinity so it makes sense that the series would not converge at x=1 because it is trying to approximate a function that is going to negative infinity.2187

We got our interval here, let me recap what we did.2202

This was the series that we derived back in example 2, that is all coming from example 2, you might want go back and check that if you have not just watched it.2205

The way we figured out the interval of convergence was we applied the ratio test, which means you look at the n + first term divided by the nth term.2216

We took our formula and we plug in n +1 for n and then we just left with n + n in place of n and immediately the negative signs disappeared because of the absolute values.2226

We sorted our fractions here and the we separated out the x's on one side and the n's on the other and the xn+1/xnand just cancels down to x the end of n +1.2239

We take the limit of that and it turns into 1, that whole terms since we are multiplying it that just all drops out turns into 1.2253

We are left with just the absolute value of x and remember the ratio test tells us that whatever you get in that limit, it converges when that limit is less than 1.2261

The absolute value of x has to be less than 1 for it to converge.2273

We know it converges from -1 to 1, diverges outside that interval we don't know what happens at the end points remember the ratio test never tells you what happens at the end points.2279

You got to check each one separately, we plug that x =1 into this series right here, plug in x=1 here and we got something that we remember from calculus is the harmonic series.2291

The negative of the harmonic series which diverges, that was something we learned back in calculus 2 and if we plug in x=-1 we get the alternating harmonic series.2305

We also learned in calculus two that converges by the test we learned from calculus two was the alternating series test so what that tells us is that it converges for x =-1 and it diverges for x=1.2319

The series converges from -1 is included up to 1 excluded and you can write that interval notation n by putting square brackets on the [-1] and round brackets on the (1), that is our interval.2334

Of course we checked because natural log 1-x the original function that we started here if you try to plug-in x=1 then you will be trying to take natural log of 0 which does not exist.2350

That is a nice check on the fact that our series did not converge at x =1, let us move on.2364

In example 4, we are going to use power series to solve the integral of e(x)2dx and this is really starting to illustrate the usefulness of power series.2373

Because this is an integral that you would not have been able to solve by the standard integration techniques in calculus 2.2389

The only way you can solve this is by some kind of approximation or by using taylor series techniques so we need to find the power series for e(x)2.2397

Now that is something that if you start to write derivatives it will get very messy, very quickly you will get tangled up in the chain rule, product rule, and the power rule.2409

It will all get very quickly involved and get very messy, instead the way to write down a power series for e(x)2 is to remember what your power series is you put to the x.2420

I hope you got that memorized, if not i recommend that you memorize it at least while you are going to be using series to solve problems.2436

But here it is, it is 1+ x + x2 2 factorial + x3/3 factorial and so on.2443

That was for ex but then you can get the series for e(x)2 just by making a little substitution.2451

Whenever I see an x, I'm going to change it to x2 and I get 1+ x2 + x(2)2, well that is x4.2457

It still over 2 factorial, that does not change, plus x(2)3, that is x6/3 factorial and so on.2468

What I really want to do is integrate both sides here, so I put my integral sign here and squeeze in a little dx where I can, dx.2480

The key point here is that now that I have converted the function into what is essentially a polynomial that goes on forever but we can think of it as a polynomial.2494

I can integrate this term by term very easily the integral of 1 is just x, integral of x is x3/3.2504

Sorry the integral of x2 is x3/3, the integral of x4/2 is x5/5 x 2 factorial.2513

The integral of x6/3 factorial is x7/7 x 3 factorial and it will keep going along that pattern.2525

Whenever I do an indefinite integral I always have to add an arbitrary constant and I'm going to add that at the beginning so that is my integral e(x)2/2, I'm sorry e(x)2dx.2536

You could also use the summation form to write this, an alternate solution would be to say that ex is the sum of xn/n factorial.2562

e(x>)2 is just the sum of x(2)n which is x2n/n factorial.2578

When you integrate that, when you take the integral of both sides, these sums all go from n=0 to infinity.2589

The integral of x2n his x2n+1/ 2n + 1 x n factorial.2602

This is another way to arrive at the same result if you expanded out that series you get that same thing that we got over here except I forgot to include my c, so I will tack that on.2621

We found our integral of e(x)2 in terms of a series which is something you never could have done in calculus 2 using the basic integration techniques there.2643

Let me remind you how we did that, I remembered my series for ex and that is much easier than trying to find the series for e(x)2 directly by writing down derivatives.2653

I just substituted in the x, i changed it to x squared, I made that substitution on both sides.2664

x(2)2 gave me x4, x(2)3gave me x6.2671

We made that substitution on both sides that converted it into essentially a polynomial, certainly in terms of integrating it i can think of it as a polynomial.2679

When I integrate both sides, my 1 turns into x, x2 goes to x2/3, x4/2 factorial goes to x5/5 x 2 factorial and so on, we just integrate it term by term.2689

I tack on the arbitrary constant and the other way to think about that is to remember that ex we have a nice σ formation for the series and e2x, sorry e(x)2.2704

We just substitute in x2 and so we get x2n in this series and the integral of that is just x2n+ 1÷ 2n + 1, that is a way of getting a series of a σ summation form for the same thing.2720

On our last example here we got y(x) is the sum from n=0 of a sub n, xn.2743

We are being given a generic power series and we want to find the power series expressions for y′ of x and y″ of x.2751

We are going to shift the indices of summation so that they start at n=0, this is actually an example that is designed to warm you up for using power series to solve differential equations which is what we are going to be using in the next lecture.2758

I really encourage you focus on how this example works because essentially the start of every problem in the next lecture is going to be exactly what we are doing in this problem.2774

It is really a key problem here but let us see how it works out if y(x) is given to us, this generic power series, that is the guess we will make to solve differential equations in the next lecture the Y′(x).2787

We are going to take the derivative there n=0 to infinity of, now I will put in a sub n because that is a constant but the derivative of xn is n x xn -1.2803

We get this new series but you will notice that the n=0 term is 0 and that is because of this factor of n right here so that n=0 term of that series really doesn't change anything.2821

In fact i can leave that off and just start the series at n=1 so we can omit that n=0 term and I can rewrite this series exact same thing except for n=1.2839

That is because instead of adding up the n=0, 1, 2, 3 terms I noticed that the n=0 term really is not doing anything, I can just add up the n=1, 2, 3, 4 terms. 2858

I'm writing the same thing here (an)x < n-1 and it is something that is shifting the indices of summation so that they start at n=0.2871

I want to show you what I mean by that if I expand out this series, I'm going to start out at n=1 and I will get a sub 1, x0 is just 1, I will not put anything there, that was from n=1 term, the n=2 term gives me 2 a sub 2 x x1 , n=3 term gives me 3(a3) x x2 and so on.2886

I'm going to forget the old version of the σ form and just look at these terms and if I want to write it in σ form now and I want to write so that there is an xn there.2917

What I noticed is these numbers on the coefficients are 1 higher than the power of x, here I got 3 a sub 3 x2 so the coefficients are 1 power lower, what I'm going to do is I'm going to put n+1.2936

I think I did leave myself quite enough space here, let me rewrite that, n + 1, a sub n + 1, xn and if I do that that all make sense if i start at n=0.2960

If you think about that if I plug in n=0 here, I will get 1 x a1 x x0 which would give me just exactly the a1 term here.2982

The n=1 term would give me 2 a sub 2x, the n=2 term would give me 3 a sub 32 and I'm going to end up getting the exact same series and I wrote above.2999

I'm just using different index of notation to keep track of that and let us look at what effectively happened here from when I expanded it out and wrote it in a new form.3013

If you look at this n=1 term, the index went down by 1 so we lowered the index by 1, because of 1 down from 1 to 0 and then all the n's in the formula, all these n's got raised by 1.3025

That n turned into an n+1, the n and the subscript term turned into n+1, and the n+1 turned into an n.3058

To fix that you raise the n's in the formula by 1, that is going to be a really useful trick for us to use in the differential equations lecture next time.3067

You really want to keep this in mind if you lower the indices by one then you raise the n's in the formula by 1 and it works the other way too.3097

If you raise the indices by one then you lower the n's in the formula by 1 and it is a way of adjusting the powers so that we can make different series be compatible with each other.3107

Let us practice that by going one more steps here by finding y″(x) so y″(x) , that is the second derivative.3119

Again I will start with n=0 to infinity now I got an xn -1, while I still have an n.3130

nx-1 if i take its derivative will give me n-1, there is still an a sub n, xn-2 , that is by taking the derivative of this expression right here.3141

I noticed that the n=0 term because of this n right here will be 0 and the n=1 term, let me write that down the n=0 term is 0.3155

Because of that n-1, the n=1 term is also 0, I can drop those two terms out and just start at n=2 to infinity of nx n-1 x a sub n/ xn -2. 3173

I'm not shifting the indices of summation yet, I'm just dropping off the first couple of terms because I noticed that they are going to be 0.3193

I'm not doing this business over here where I'm shifting indices that is what I'm going to do next.3201

I looked at this and I see that xn -2 right there, I want to raise that up and make it xn.3206

What I'm going to do is I'm going to raise the n's in the formula by two in order to get that n -2 up to an n.3214

To make that work I have to lower the index by 2, by the same principle, I will not bother writing out the series this time, it is the same principle as before, so that n=2 in the index, I'm going to drop that down to n=0 and I'm going to raise all of these n's by 2.3237

I have n turned into n +2, n-1 raise it by 2 and gets it to n+1, a sub n becomes a sub n+2, xn-2 becomes xn.3265

Let me box what I have figured out, your Y′ is n= 0 to infinity of n+ 1, (an) + 1xn and Y″ is n= 0 to infinity and +2 x n +1 x (an) +2 x xn.3281

These are really important formulas, it is worth studying them very carefully and making sure you understand them or at least to remember them.3304

Because we are going to be using them very heavily next time when we use series to solve differential equations.3312

Let me recap how we obtained those formulas, we started with y′ of x I took the derivative of this so the derivative of xn is nxn-1 .3319

I noticed at the first term that n would just be 0 anyways, I dropped it out and I started at n=1, instead at n=0 and dropping off the n=0 term.3333

I did this shifting index trick where I lowered that one down to 0 and then I raised all these n's and the way we can justify that if you want to is to expand out both this series and that series.3344

If you expand out either one you will see that you will get this series either way, those are really the same thing.3359

That was my formula for y″, I took another derivative of xn -1 and I got n-1, xn-2.3367

I noticed that since I have an n and an n-1, the n=0 and the n=1 term are both going to be 0 so I drop those out and I can start my index at n=2.3381

That was not shifting the index yet that was just dropping off a couple of 0 terms and I'm going to use the same principle where I will lower the index by 2, that is why that 2 became a 0.3394

Then I raise each of these n's by 2, so that n became n +2, that n -1 became n +1, that little n became an n + 2, and that n - 2 became an n.3406

If you do not trust that, if that seems like magic, just expand out each one of these series, expand out this one and expand out this one right up the first few terms of each one.3423

You will see that you will get the same series either way and this is going to be absolutely crucial for us when we use differential equations to solve our new series to solve differential equations at the topic of the next lecture.3433

And that concludes our review of power series for this lecture.3448

You have been watching the differential equations lectures on www.educator.com. My name is Will Murray, thanks for watching3452

Hello and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going to talk about series solutions of differential equations.0000

In the previous lecture we did a review of power series, if you are a little rusty on power series and you have not worked through that stuff yet, it might be worth looking back at that previous lecture before you jump in to this one.0010

Because this one is a pretty heavy one especially it might be worth looking back at example 5 of the previous lecture where we learned how to take the derivative and the second derivative of a power series.0021

You can adjust the indices and the exponents to make them fit into a differential equation.0035

We are going to be using that idea very heavily in today's lecture on how to use series to solve differential equations.0041

Let us go ahead and get started here, the idea is fairly simple here, it is just the mechanics that are complicated, there is no difficult theory here but the mechanics get to be quite tedious and you got to be pretty careful here.0048

The ideas that you guess a power series solution to a differential equation and then you plug it in and in order to plug it in, you got to calculate its derivatives.0062

You start out with this generic power series y(x)= the sum of n=0 to infinity of a sub n x xn and the derivative of xn is just n x xn-1.0071

Because that first term, that n=0 term is going be 0, you can drop it off, which means you can take the index and you can start right at n=1.0091

We dropped off the n=0 term and if you expand this series out, plug in n=1, you will just get a1, n=2 gives you 2a/2x, n=3 gives you 3a/3x2 and so on.0101

You can write a new formula for this where you start from n=0 and you have n +1 x a/n +1 x xn, that is still the same series.0119

If you expand out this series, you will still get the same thing here but it is a different formula for it and remember the mnemonic to keep track of that is that we lower the index.0134

We lower the index by, in this case by 1 and we raise the n's in the formula, raise the n's, this is going to be really useful.0155

It is very important to remember how this works so we lowered this index from n =1 to n = 0, and to balance that out we raise these n's.0168

These n became an n+1, this n became an n +1, and this n -1 became an xn, we would be using in every single example today.0178

We will lower the index, we will raise the n's in the formula, let us keep going and we also need to look at the second derivative.0189

The second derivative, remember we had n/a sub n, xn-1 for the first derivative.0198

For the first derivative, when we take the second derivative of xn -1, we get an n-1 x xn - 2 and everything else comes along to the right.0208

If you noticed here the n =0 term is going to be 0, and the n=1 term is going to be 0, both of those terms have no effect on the series so we can drop those out and start the series at n=2.0221

This is the exact same thing if we just start the series at n=2 because those first two terms really have no effect on it, we are going to use that same trick where we lower the index.0238

We lower the index by 2 and to balance that out we take all the n's in the formula and we raise them by 2.0254

Plus 2 here, n turns into n + 2, n - 1 we raised it by 2 terms into n +1, the n turns into n + 2, and the n - 2 when we raise it by goes up to n.0260

It is that same trick of lowering the index by a certain amount and raising all the n's in the formula by a certain amount.0279

The reason we want to do this is we want everything to match up at the powers of xn , will be doing a lot of adjusting these powers in order to make things match when we solve differential equations later on.0286

Let me tell you what happens when you plug these things in, you got to plug these series into the differential equation that is given.0300

What you want to do is match them up to each other and it is really a 2-step process.0307

First step here is to match the exponents on x, what that means is you might have some series that are in terms of xn -1.0311

Some series that are in terms of xn maybe some other term in terms of xn + 1.0320

You have always different exponents on x, first you want to try to get them all to the xn, you want to change all these into xn.0327

The way you do that is by doing that index shifting trick that I just though you in the previous side.0337

The first thing that you do is try to get everything to be equal to xn.0346

The second step here is you might have some series that started at n=0, some series that starts out at n=1, and some series that starts out at n=2.0351

You want them all to start at the same place and the way you do that is you pull out terms from the ones that we started earlier.0366

For example in this case, if we had an n=0 series, we write that as we pull out a couple of terms and so we can start at n=2, n=0 we would write that as we just pull out the n=0 term, pull out the n=1 term.0375

Wherever that n=1 term is and then you could start the same series at n=2, if you have an n=1 series, you could pull off the n=1 term and then you could start at n=2.0397

What you want do is after you got the the exponents to match up, then you want to pull off the starting terms so that all the series starts at the same place.0417

We are going to find the recurrence relation on the coefficients by setting the coefficients equal to 0.0430

That is not so obvious right now until we do some examples so do not worry about that if that does not make a whole lot of sense, it will all start to make sense as we do a few examples.0435

We really need practice with this, we are going to use the recurrence relation to solve for higher coefficients in terms of lower ones and then we use our coefficients to build our series and to create our solutions.0444

A lot of this, I think probably will make sense until we really work on some examples so I think the best thing to do is just jump in right away and try to solve an example here.0456

First example, we are going to guess series solution to the differential equation Y′ - 3x2 y = 0.0470

We are going to plug in our solution and try to find the recurrence relation on the coefficients.0477

We always start the same way with these things, we always start with Y = the sum a/n, xn, sum from x=0.0484

All of these series go infinity, I'm not even going to bother writing the infinity on top of the series.0496

What is important is where they start that might be changing, so Y′= the sum from n=0, of n x a/n, xn - 1.0502

We saw in the introductory slide that we can rewrite that as the sum on n = 1 of n(a/n) xn- 1. 0519

That is because the n=0 term really does not do anything but now I'm going to look at the differential equation and I see that I'm going to be looking at 3x2y.0540

I'm going to figure out what 3x2 y is.0548

3x2y, I look up at my Y series that is up here and it is the sum from n=0 of 3 x a/n and now I had an xn but x2 is going to bump that up to xn +2.0552

These are the 2 series that I'm going to be trying to combine, this xn -1 right here and this xn +2 right here and remember there is a two-step process here.0570

The first step is to match exponents, it is very important that you do this in order by the way.0583

First step is to match exponents at xn and the second step is to match starting indices0591

That is the place where your starting the n for each one of these series so I look at these two series and I see I got one starting with xn -1 or 1x - 1, 1x + 2.0608

I got to reconcile all those, I got to match them both at xn and the way I do that, I have to raise this n by 1 which means I'm going to lower the starting index by 1, that is going to be n=0, n + , a/n + 1.0623

Remember that trick we used to raise all the n's in the formula by 1 and you lower the index by 1 so that is really that trick right there.0644

I do -1 on the starting index and +1 on all the n's on the formula so all those n's in the formula got a +1, got a-1 on the starting index.0655

A very common trick, we are going to use it over and over and over again.0671

Down here, the 3x2y, I see an xn + 2 and I want to match that up with xn.0675

That means I have to lower those n's in the formula so I have to lower them by 2, 3a sub n - 2, xn and the price of that is I have to raise the starting index by 2.0684

I will start with that at n=2, whatever you do to the index, you do the opposite to the formula, there I went + 2 in the index and the n's on the formula each one did got a -2,that is nice.0702

What I have done here is I got an xn and on both of these series, I have matched up my exponents, I have done that part.0723

I will make that green to show that I have done something good, I have matched up xn on both series but now I see the 1 series starts at n=0 and the other series starts at n=2.0732

I need to match my starting indices, that is the second step of resolving these 2 series, to match the starting indices, what I'm going to do is take the lower n=0, and I'm going to pull out a couple of terms.0747

I'm going to pull out the n=0 term and the n=1 term out of that series, then I can start it at n =2.0762

If I start at n=0, n=0 would be n +1 would be 1, that would just be a1 x x0, n=1 term would be to 2a/2x and then I would just have the rest of the series so I can start it at 2 and n=2.0771

n + 1, a/n + 1, xn, turns that right a little more neatly for you.0796

That is nice because what I have done there is I have matched that starting index with that starting index, they both started at n=2.0811

I have achieved my second goal here of matching the starting indices and now I can plug these 2 into the differential equation.0820

I got my Y′, this one was Y′ and this one was still 3x2y, I'm going to plug those into the differential equation Y′ is a1 + 2/2x + the sum from n=2 of n +1, a/n + 1, xn.0828

Now, - 3x2y - the sum from n=2 to infinity of 3a/n -2xn and that is equal to 0.0861

I'm going to gather my terms together I got a 1 + 2a/2x and I'm going to combine these since they started at the same place and the same power of x.0876

That is what I did all this work up here for, n=2 of n + 1 x a/n + 1 - 3a/n -2, all of that times xn =0.0887

The way you want to think about this is really like a polynomial, I got my constant term over here at a1, my x term here at 2a/2 and then these are the higher powers, the polynomial, this kicks in at n=2.0909

This gives me my x2 term, my x3 term, and so on.0924

I want to think about this as a polynomial on both sides and I want to equate coefficients on both sides of the polynomial so my constant term on the left hand side is a1.0928

On the right hand side it is 0, my x term on the left hand side is 2a/2 and on the right hand side it is 0.0945

My x n term where for our higher powers of x on the left hand side I have this big formula n +1 x a/n + 1 - 3a/n - 2 = 0.0958

I know that it is valid for all n's bigger than 2 and it is coming from right there, this is true for n bigger than or equal to 2.0980

That is called the recurrence relation and that is what we are going to use to solve and find higher coefficients in terms of lower coefficients.0994

That is the recurrence relation right there, that is what always the problems are asking for is to find the recurrence relation and that we just found right there.1005

Let me recap quickly how we did everything here, we started with our generic power series for y, we wrote down Y′ and then we drop out the n=0 term.1017

That is why we started at n = 1 and then because we have to plug this in to the differential equation.1028

I was looking at 3x2y, I took my Y term multiply it by 3x2 and that is why I got xn + 2 here and I wanted to match my exponents xn on each one.1036

In order to make that turn into xn, I had raised all my n's here, the price of that was lowering my n here in order to get this want to be xn I had to lower my n here, I have to lower all the n's in the formula, I had to raise all the n's in the index there.1052

I get all my n's matching up the exponents, then I had to match my starting terms and I saw that I had n=0 on one and n=2 on the other.1073

In order to make resolve that, I take the n=0,1 and I pull out a couple of terms, I pull out the n=0 term, pull out the n=1 term and I get the same series starting with n=2 which is good because it matches this series.1084

I plug them both into the differential equation that I'm actually trying to solve so I get this is all Y′, this is -3x2y =0 and I combine my 2 series and then I think of this as a big polynomial.1100

I got a constant term equal to 0 an x term equal 0 and then this is the coefficient of all my terms for x2 on up, that is the recurrence relation and is valid for n greater than or equal to 2.1124

We are going to keep on working on this example, we have not finished solving this differential equation yet, we are going to use this recurrence relation to build our solutions.1139

That is what is coming up in the next example, in example 2 we are going to use the recurrence relation that we derived back in example 1 to solve Y′ - 3x2y =0.1148

Let me remind you all of this stuff up here was the stuff we figured out in example 1, if are just tuning in and have not looked at example 1 recently go back and check your example 1.1161

That is where all this material comes from this business about a1=0, 2a/2=0, and so on, that is all coming from example 1.1175

What we are going to do with that is to build our recurrence relation and solve for higher terms, in terms of lower terms.1188

What I'm going to do is take this and I'm always going to solve for the higher one in terms of a lower one, the higher one is a/n +1, the lower one is a/n -2.1200

I get n +1 and I'm trying to solve for a/n + 1, move the the other one over to the other side.1211

That is 3 a/n - 2 and then I will solve for an +1 is 3a/n -2/ n+1, and whole point of this is that now I can plug in smaller values of n and I can find larger coefficients in terms of the lower ones.1220

Remember this is valid for n bigger than or equal to 2 and I'm going to start plugging this in with n bigger than or equal to 2.1244

I will start out with n=2 and that will tell me that a3 because it is a/n + 1 is equal to 3 a/n -2 so 3(a0)/n + 1 is 3 because n=2 so that is just a0.1251

We already figured out a couple of terms already the a1 and the a2 we do not know what a0 is.1272

I will just write up here a0 is arbitrary, we have not figured out anything about a0 but we know a1 and a2 and we are starting to figure out the higher coefficients.1281

We figured out a3 right here in terms of a0, our next value to plug in is n=3, that will tell me that a4 is equal to 3 x a, well if n=3, it will be 3a1/4 but I figured out that a1(0) here.1293

That will be just 0, n=4 will give me a5 in terms of 3(a2)/5.1314

I'm using this recurrence relation over and over again, but a2 were 0, that is what we figured out up here so that is still zero, n=5 will give me a6 will be 3, now if n=5 it will be a3 /6 which is (a3)/2 but a3 was a0.1324

That is (a0)/2, it is starting to build up higher coefficients in terms of the lower ones, n=6 will give a7 in terms of a4.1360

It will be some multiple of a4, I do not know exactly which one but I do not really care because I already figured out that a4 is 0, n = 7 will give me a8 in terms of,1376

Let me make that more clear, that was something I did not bother to figure out, a8 is in terms of a5 and I do not really care because it is going to be 0 anyway because a5 was 0.1395

n=8 will give me a9 will be 3(a6)/9 which is (a6)/3, but remember (a6) was (a0)/2 so it is a0/2 x 3 and we are starting to figure out this pattern.1406

We get 2 zeros and then something interesting, a10 will be 0, a11 will be 0 , a12 will be a0/2×3 x 4 and what we clearly see is that we see a factorial pattern building up on every coefficient.1443

(a3)k or 3n will be a0/n factorial because a12 was a0/4 factorial.1464

Let us write down the series that this generates, remember our original Taylor series was y is equal to the sum of (an)xn and starting from n=0.1478

This is a0 + a1(x) + a2(x2) + a3(x3) + a4(x4), and so on.1494

Most of these coefficients are actually 0, a0 we do not know what that is, remember we say a0 was arbitrary.1508

ai is 0, a2 is 0, a3 is a0, the next non zero one is going to be a6 which is the coefficient of x6 + a0/2 x6 + a0/2 x 3.1518

That is 3 factorial x x9 + a0 /4 factorial x x12 and so on.1548

What we get here is, if we write this in closed form. I am going to factor out an a0, the sum from n=0 to infinity, we still have powers going up by 3 0's, I'm going to write x3n/n factorial, remember 0 factorial is 1.1563

If you plug in n=0 here, you will get a0 and I see that I forgot my power of x3 on this term right here.1585

This is x(a)x3 that is coming from that term right there, if you plug in n=1 you will get x3/ 1 factorial.1595

That is where that is coming from and if you plug in n=2 you x6/ 2 factorial and in general we get x3n/n factorial.1604

I remember the Taylor series patterns that we reviewed in the past lecture, you might want to check them out in the past lectures if you are a little rusty on your Taylor series.1616

I recognize this one because I remember ex is 1 + x + x2/2 + x3/ 3 factorial and so on.1625

The short version of that is xn/ n factorial, what I got here is not quite that, I get x3n but that is the same as that a0 is just my arbitrary constant, I will call it c.1638

(c)(x)3 because I have the formula for ex except it is x3n, I will just substitute it in x3 in place of x.1654

Finally, we got a solution to our differential equation, let me recap where the different elements came from there.1661

These initial expressions all came from example 1, go back and look at example 1 if these are still mysterious to you.1678

What we did was we took this recurrence relation and we solve for the higher term that is the n +1 term in terms of the lower one which is the n -2 term.1687

That is what we got here, where we solved for n + 1 + a/n + 1 in term of a/n -2 and that was valid for n being bigger than 2.1698

Back in example 1, that was because our sums started at n=2 that is where that 2 came from and we started plugging in different values of n into this recurrence relation starting at n=2.1707

Each one gave us a higher coefficient in terms of a lower coefficient and a lot of these coefficients turn out to be 0 because of these 2 starting coefficients being 0 here.1721

But we did not have an expression for 0, we had to leave the one that is in terms of a0 and we got this pattern building up and it turned out to be this pattern of factorials.1733

When we plug in back into our original Taylor series this was our original guess and all the 0 terms dropped out.1747

But a few terms we are left and we built up this factorial pattern in the series and then we recognized that, that is actually just like the series for ex except that that x has been replaced by x3.1757

What we have here is x^e(x^3), the c is just a different letter for a0, if you do not like the c just call it a0 it is just fine.1773

What we end up with is a constant x e^(x^3), we will do more examples and we will get more practice with this.1783

Let us keep moving now, we are going to guess a series solution to the differential equation Y″ - XY′ - y=0.1793

We are going to plug it in and we are going to find the recurrence relation on the coefficients.1807

This is going to work the same as the other one, we start with the same guess every time and you will get used to it after a while.1813

Y is the sum from x=0 of an(x^n), let me make this a little faster this time because it is these initial parts is the same every time.1819

Y′ is the sum of n x a's of an x^n - 1, just taking the derivative and we noticed that the n = 0 term is 0.1830

I'm going to start this right away at n = 1 and then Y″ is the sum of n x n - 1, a's of n, x^n - 2 and the n=0 term, n = 1 term are both 0.1842

You can start at n=2, if you look back at the beginning of the lecture in the lesson overview, I went through this a little more slowly.1862

If you want a little more practice and it is not quite making sense yet, what we want here is we want to plug these into the differential equation.1869

We want to figure out what XY′ is, let me write down my XY′ is, well that is just pumping it up by power of x, that would be the sum of n=1 of n^a(sub)n.1880

Now x^n - 1 is going to become x^n because I'm pumping it up by a power of x, there are 2 steps to getting these things all to be compatible with each other.1896

I will write them up here, we talked about this at the beginning of the lecture when we first matched the exponents.1908

x^n, we want everything to match in the exponents and secondly we want to match indices.1921

That is the starting indices where each series starts there, there are 2 steps to that and you got to do that in order.1933

You do not want to try that, it makes up the order at the road to doom and despair, you want to look at these exponents first.1939

What I see here is, I see an n here that looks pretty good, I see an n here that looks nice but this is an n -2 and I want to match that up to be an n.1948

That means I want to raise the exponents here and that means I have to lower the exponents.1958

Let me write this in red, I want to do a + 2 here to make that work and the price of that would be going - 2 down here so this is the same as the series from n = 0.1967

I have to do + 2 n all the n's in the formula, that would be an n + 2 x n + 1, because n -1 + 2 is n + 1, a/n + 2 x^n.1983

I got an x^ n everywhere, I have matched my exponents and I got a nice x^n there, and there are very nice x^n there, and there are beautiful x^n there.2001

The problem here is that I started to match my indices , I look at my indices and I see n=0 here, n=1 here and n=0 here.2012

That means I'm going to pull terms out of the lower ones in order to make a match and it looks like I can make a match on n = 1.2024

I'm going to pull out n =0 term out of this first one here, the n = 0 term would just be a0 and then I can restart the series at n=1, n=1 ax^n and here I need to pull out the n = 0 term.2032

The n = 0 term would be to 2^a2 and then I can restart my series at x^n at and n = 0, n + 2, n + 1, a/n + 2x^n.2052

I got each one of those series now starting at n = 1, that is very nice that those all match and of course the exponents still match so I'm ready to combine this together.2073

I'm ready to plug them all into the differential equation, I see my Y″ and I'm going to take the 2^a2.2089

I'm going to go ahead and combine as I plug them in so my Y″ is n=1 of n + 2 x n + 1, a's of n + 2 and that would be an x^n here.2101

That is my Y″ term , now I see a -XY′ so that is this term right here so there is no separate terms on this one.2125

I'm just going to minus a and n, and then there is a -y right here, that is -y is -a0 - a/n x^n.2136

That whole thing is equal to 0 so I was just plugging each one of my expressions into the differential equation and let me clean up my starting index here because that is very important.2154

Those are starting at n=1 again I want think about this as a big polynomial, I think about this as 0 + x + 0/x + 0/x^2 on the right.2168

On the left, I look at this and I see that this is my constant term and so my constant term must be equal to 0, 2^a2 - a0 = 0 which I can solve that out for a2 always solve for the bigger one in terms of the smaller one.2182

I get a2 is equal to a0/2 and then this is my recurrence relation for the higher power starting from n=1.2204

I'm going to solve that out n + 2 x n + 1 x a/n + 2 -, 2217

I'm going to factor out the a(sub)n - n +1 times a(sub)n =0 and that is all for n greater than or equal to 1 and the place where I got that one.2228

Let me highlight that, that one came from that one right there so I always solve this for the bigger one in terms of the smaller one.2241

Here the bigger one is a/n + 2 so I get to and we can cancel n + 1 from both sides divide that away very quick and then I will get n + 2, If I move the (an) over the other side and divide by n + 2, I get (an) +2 is (an)/n +2 and for n bigger than or equal to one.2251

That is my recurrence relation that is going to tell me how to find bigger coefficients, bigger (an) in terms of smaller ones and that is the starting one is equal to a0/2.2280

That is the end of that example but let me remind you how that worked out rather quickly and the reason what I'm going through a little more work quickly now is that the starting steps are the same in all of these.2298

They all start out with y=(an)x^n, Y′ is equal to n(an) x^n -1, Y″ is n x n -1an x^n -2.2310

Those are the same in every single series problem ever and is also true that you can drop off the n=0 term here and start it at n=1, drop off the n=0 and n=2 term, I'm sorry n=0 and n= term from Y″ and start it at n=2.2323

Those are always the same in every series problem then what you do is you start to look at the differential equation itself and you start making modifications.2341

In this case, I looked at this differential equation and I saw I have an x x Y′ that is why I figure out what XY′ is here and that bumps my power of n -1 up to n here.2351

I start trying to make the series compatible with each other so I'm going to match the exponents first got an x^n here, x^n here, here I did not have an x^n.2365

I wanted to raise these n's by 2, raise all the n's in the formula by 2, the price of that was to lower the index n by 2.2381

I match my exponents, but then I want to match my indices so I see, the starting index here is zero, here I had 1, in here I had 0.2389

In order to bring those zeros up to 1, I pulled off the n=0 term on each of these series and then I can start each one of these series at n=1.2403

In this case, we are everything ended up starting at n=1, once I got all the series compatible with each other meaning that the same exponents the same starting indices I can plug them into the differential equation. 2414

That is what I was doing here, is I was plugging everything into the differential equation and I sorted things out as I did it, I sorted out these extra constant terms on the outside.2427

I combined all the x^n terms on the inside, I set them equal to 0, that is from right here and then I read this like a polynomial which tells me that my constant coefficient must be zero, I can solve for a2 in terms of a0.2440

And more importantly this recurrence relation right here must be zero and that is what I did here and that allows me to solve for an + 2 in terms of (an). 2458

You always solve for the bigger coefficients in terms of the lower coefficients, I will solve for a2 in terms of (an) and it is true for bigger than or equal one, that is coming from that one right there.2471

That is the recurrence relation for this differential equation, we still have not solve this.2485

What we are going to do in the next example is take this recurrence relation and work out what the coefficients must be starting from the smaller coefficients2489

You want remember this formula right here and this one right here, those are the one we are going to be using in the next example.2498

On example 4, we are going to use the recurrence relation that we derive above in example 3 to solve this differential equation, so let me remind you what recurrence relation we got here.2508

We got it right at the end of example 3, everything here is coming from example 3, if you have not just watched example 3, you might want to go back and watch example 3.2524

That you know where are we getting these formulas because otherwise that would be a complete mystery.2537

Basically, what we figured out in example 3 with this came from the constant term of the differential equation and we figured out from this the a2=a0/2.2542

Then we figured out from this long recurrence relation, we figured out that (an) +2 and let me write that up here because I'm going to be needing some space down below.2557

(an) + 2 we solve this down, we figured out that ,that is in terms of an/n +2 and that is valid for all n bigger than or equal to one, I have not really done any new Math this example yet.2570

All of this is coming from example 3, you can check back and see where this comes from, let me show you where were going with this.2584

We are going to find out the coefficients a0 I got nothing here that tells me what a0 is, a0 is arbitrary and I got nothing that tells me what a1 is.2593

a1 is also arbitrary, nothing here will tell me what a0 and a1 are, a2 I figured it out right here in terms of a0, I'm okay for a2 and now I'm going to start using the recurrence relation starting with n=1 because that is what it is valid for.2605

If I plug in n=1 to the recurrence relation I get a3 that is what I'm looking right here now, a3 is equal to a 1/3 and if I plug in n=2, I get a4 is equal to a 2/4 and remember I already knew my a2 in terms of a0 so this is a0/2 x 4.2632

plug-in n=3 to the recurrence relation I get my a5 is equal to a3/5 but my a3 I knew in terms of a1, that was up here so this is a1/3 x 5, n=4 gives me a6 is a4/6, that is a0/2 x 4 x 6.2666

We are starting to see a pattern here, I see that the even ones are building up even numbers in the denominator and the odd ones are building up odd numbers in the denominator. 2697

I will write down one more term, n=5 will give me a7 is a5/7 which is a5 was a1/ x 5, so this all comes back a1/3 x 5 x 7 and now I think I'm ready to write my solution for the differential equation.2708

My original guess for y was the sum of an/x^n and and then we just expand that out and started at n=0 so that is a0 + a1x + a2x^2 + a3x^3 + a4x^4 and so on.2733

I'm going to plug in what I know about those coefficients, a0 I do not know anything, I'm just going to leave that as a0, same with a1, I have to leave that as a1x, but now a2, I figured out that ,that is a0/2.2760

I'm going to plug that in as a0/2x^2, a3 was a1/3x^3, a4 ws a0/2 x 4x^4, a5 was a1/3 x 5x^5, a6 was a0/2 x 4 x 6x^6, a7 was a1/3 x 5 x 7x^7 and so on.2775

What you see here is everything got either an a0 or an a1 and I think I want to separate out my a0 terms and my a1 terms.2821

My a0 term is from here I got a 1+ x^2/2 + x^4/2 x 4 + x^6/2 x4 x and so on.2833

I will segregate out my a1 terms and I see I got an x + x^3/3 + x^5/3 x 5 + x^7/3 x 5 x 7 and so on.2859

What I really done here is I built up 2 independent solutions, you want to think of the a0 and a1 as the two arbitrary constants just like the C1 and C2 that we had in earlier solutions for differential equations.2880

You think of this is an arbitrary constant x one solution and an arbitrary constant x another solution.2895

There is a few clever algebraic tricks that we can do with the solutions that will help us recognise them as something that we have seen before.2904

In particular, for example this 2 x 4 x 6, I can write that as 1 x 2 x 2 x 2 x 3 x 2 and then I can write that in turn I can pull out a 1, 2, 3, x 2 x 3 x 2 x 2 x 2 and that is 3 factorial x 2^3.2914

We are going to have a similar pattern on every term, in that first solution this is a0 times, it looks like I'm getting even powers, I'm going to write x^2n and in each one of these, I'm going to have an n factorial and a 2^n.2940

That goes to n=0 to infinity, in the right hand side it looks like I'm building up a pattern of odd numbers in the denominator, that does not factor out so nicely, I will just write 1 x 3 x 5 up to 2n +1.2960

I have an x^ 2n + 1 in the numerator, again starting from n=0 because we are using 2n + 1 to the first term term what we will get is 1, a1 there.2983

There is something else you can do that that is kind of clever here which is to multiply on the even numbers to fill in the factorial on the bottom.2998

If I write 2 x 4 x 6 up to 2n and I balance that out with 2 x 4 x 6 up to 2n on the top, what I will get in the bottom is 2n + 1 factorial because I filled all the even numbers to make a nice factorial.3008

In the top, I saw I have x^2n +1 and that 2 x 4 x 6 up to 2n, remember that is the same thing I was dealing with over here, I can factor that out as 2^n x n factorial.3030

That is not absolutely necessary but it cleans up our product to make it a little nicer here, for my second series, I will write a1 times the sum from n=0 to infinity of 2^n x n factorial x x^2n + 1/2 + 1 factorial.3046

That is a slightly cleaner way of writing it, that is not absolutely necessary but it is a little nicer, what I noticed on the left, this x^2n/2^n , I can write that as x^n(^2)/2 raise to the n power and then I recognise a formula here.3073

I got something to the n/n factorial and here is what I remember, e^x remember is the same as x^n/ n factorial, what I have over here is x^2/2 e^x(^2)/2.3092

Because if you substitute in x^2/2 to this e^x formula, it is exactly what you will get right here.3118

My a0 and a1 those are just arbitrary constants, to make them look more like our solutions from before, I will write that as c1 and c2, or you can leave them as a0 and a1, there is nothing wrong with.3126

There is nothing so clever I can do with these series on the right, I can not resolve that into anything I recognise, I'm just going it the way it is, 2^n x n factorial, x^ n + 1 and then 2n + 1 factorial.3140

That is my ultimate solution there, let me box that off and we noticed here that we get two different types of solutions here.3159

One is a nice elementary function e^x(^2)/2 and the other one is this kind of are ugly power series but it's the best we can do, we can not resolve it into an elementary functions there.3174

Let me just recap what we done here, are we started out with the information that we figured out in example 3 that was the a2 we found that in terms of a0.3186

Then we had this crucial recurrence relation where we found an +2 in terms of an.3199

That was valid for (n) bigger than or equal to 1, that was all coming from example 3, we did not figured out any of that right now.3207

Based on that information, while there is nothing there tells me what a0 is or what a1, I have to leave those as arbitrary constants but once I established those values then I can figure out all the higher ones in terms of lower ones.3213

This told me a2 in terms of a0 and by plugging in different values of n into this recurrence relation, I found values of a3 in terms of a1, a4 in terms of a2 which goes back to a0.3229

a5 in terms of a3, which goes back to way a1, a6 in terms of a4 which goes back to a2, which goes back to a0, a7 in terms of a5 which goes back to a 3 which goes back to a1.3245

Basically everything comes back to a0 and a1 and when I write out my original guess for y this large series everything turns into a multiple of a0 or a1.3259

That is really nice and I can factor things out in terms of the a0 stuff and all the terms that have factors of a1.3274

After that down is a matter of algebraic cleverness and sort of recognizing the patterns of these coefficients.3285

By the way you really do not want multiply these coefficients together, you want to leave them factored instead of writing eight here and what is left is 2 x 4 x6.3292

You will notice know that there is a pattern if you leave it factored, and that is what happened here, 2x 4 x 6 I could expand that out as 1 x 2 x 2 x 2 x 3 x 2.3303

I will get three powers of 2 and I get a three factorial so I get this sum 2^n power and the factor and this factorial factor.3315

In the numerator I get at even powers of two, I have x^2n but then I recognize that is x^2/2 raise to the n so that looks like my Maclaurin series that I memorised for e^x and that turns into e^x(^2)/2.3327

On the right, this series with a1 does work out nicely I get these odd powers, these odd numbers building up in the denominator and so in order to clean that up a little bit.3347

I do not actually have to do but it makes look nicer, I multiplied it on the even powers and then to balance that I had to multiply them on to the numerator as well.3357

In the denominator, I got these factorial in the numerator using the same idea I wrote the even powers as 2^n x n factorial and that is how I got this second independent solution here in the second series.3367

There is no way you can resolve it into an elementary functions just can not leave it as a series and then my a0 and a1 where our arbitrary constants.3384

I just chose to call them c1 and c2 so they would look more like our solutions to the earlier differential equations, whenever we had second order differential equation it was always c1 and c2.3394

a0 and a1 played the same role the arbitrary constants here and I just relabelled c1 and c2 to remind you that it has the same format as our solutions to the earlier differential equations.3408

We are going to try another one, it is pretty similar to this one, you might want to try it on your own if you feel up to it and see how you do it and and now maybe try watching the video.3424

Watch me solve it with you and us see if your answers agree with mine.3435

Let us go ahead and take a look at that, example 5 we have the differential equation Y″ -3 XY′ -3y=0 and we want to find a recurrence relation on the coefficients.3440

This one is going to work out a lot like the previous one in particular it is going to start out very similarly, let me start that out my Y guess is always the same place for these differential equations.3464

Y is equal to the sum of n=0 of (an) x^n Y′ and the n=0 term drops out, we get n=1 n(an)x^n-1, Y″ is equal to n=0, n=1 term they both drop out, we can start at n=2, n x n -1 x (an)x^n - 2. 3476

That part is the same for all differential equations when you are using series to solve them.3508

The next part is where it starts to vary where you look at the differential equation and you start planning to plug these things in.3514

Then I'm going to look at this first one and I see that I have a 3x Y′ so we are going to multiply on 3x here, 3xY′ is still the sum from n=1, 3n/an.3523

Now the x bumps up that power of n -1 up to x^n and I see would not have to do anything else two major here.3543

But I want to make all these series compatible with each, remember that it is a two-step process, you always want to match exponents first at x^n and then you want to match the starting indices.3554

That is the the two steps that you want to go through and you want to make sure to do those in order every time you solve one of these things. 3579

I'm looking at my exponents, I got a nice x^n there, down here all I see I got an n -2, I'm going to bump that up to x^n which means I'm going to be adding 2^n's to the formula.3586

Let me show right here, I want to go +2 there which means I need to subtract 2 from the n in the index.3602

I'm going to start that at n=0 because I'm subtracting 2 there and then I have to add 2 to all the n's in the formula.3611

I'm going to do n + 2 x n + 1, I'm adding 2 to all these n's, (an) + 2 and x^n here because I added 2 to all those, that is good.3617

What I have done here is I got an x^n and all of the exponents here, I achieved that first step of matching the exponents, the next step is to match the indices and let us see what we have here.3634

Here I have an n=0, here I have an n=1, and here I have an n=0, now you always want to match at the higher one by pulling out terms from the lower one's so the higher one here is n=1.3649

That means I have to pull out terms from the lower one's, I'm going to write my n=0 term separately here, that is a0 and then I can start it at n=1 (an) x^n.3663

Here, I have to pull out at n=0 term, I think I'm going to do that on the next line here.3681

That is my n=0 term is 2 x 1 x a(sub)2 x x^0 and that was from n=0 and then I'll start the series at n=1 same series I just pulled off the beginning term there.3690

n=2, n + 1, (an) + 2x^n and now I want to combine this together in the differential equation, I see I have left a′ here that was Y″ that I was working on right there so make sure you make a note of that.3714

Now, I want to combine these into any one combined series so I'm going to put them together according to the differential equation.3736

My Y″ is 2(a2) and then I got a plus from n + 1, n+ 2, n + 1, (an) + 2 and then we will have an x^n here.3748

I will give myself a little more space because I think I'm going to have some more terms coming in there, -3x, -3xY′, minus this term there is no extra outside terms there so I will just do -3n(an) -3y.3769

That is -3 times all of this stuff here, I'm going to separate out -3(a0) there and now -3(an) and all of that is being multiplied by x^n.3790

Let me give myself a little more space in here, this was the sum starting from n=1 of all this stuff multiplied by x^n and that is all equal to zero.3811

What we can say from this is that we think of this as a polynomial until each coefficient has to be zero, what we get from this here is my constant term.3827

Those are constants so 2(a2) - 3(a0) is equal to 0 and my recurrence relation that I'm going to be using from that is coming from this series term.3843

My recurrence relation comes from, I think I'm going to need a little more space so I'm going to work up here, n + 2 x n +1 (an) + 2 -3n(an) - 3an=0.3863

That is true for whatever values of n you are summing over, If I go back and I look here that summation started at n=1.3890

This was for n greater than or equal to 1, that is the recurrence relation that I'm working with , I'm going to clean that up a little bit.3901

I can see that I can factor here I can write -3(an) and then there is n + 1 there, that is going to be equal to 0 and since I have a factor of n +1 everywhere I can divide both sides by n +1.3908

I cancel those off, I get n + 2 - 3(an) is equal to zero and if I move the 3(an) over the other side, it becomes positive now.3926

an + 2 is equal to 3(an) over n + 2 and that is my recurrence relation, remember you always want to solve for the higher coefficient in terms of the lower one.3943

Here the higher one is this n +2 and the lower one is (an), that is why we are solving for (an) + 2 in terms of (an) and now this constant term will solve for the higher in terms of the lower one.3958

That would be a2 in terms of a0 that will be 3(a0)/2 is the relationship I get from that. these are the 2 expressions that I'm going to use to figure out my coefficients.3973

I'm going to figure out whatever I can from these two expressions and that actually is going to carry over into the next example.3989

Just keep an eye on these two expressions, I will use them in the next example but let me quickly recap what we did here.3996

We started out with the same y, Y′, Y″ as for every other problem, those always the same, a=0, anx^n, n=1, n x (an)x^-1. 4003

That is because the n=0 term drops out because it is zero, n=2, n x 2 x n -, (an)x -2 that is because the n=1 and n=0 term dropout.4014

Then we look up at the differential equation and we see we have to multiply the Y′ by 3x, that is what I did here was I multiply the Y′ by 3x which bumped the power of x up by 1 from n -1 to n.4030

When we went up, we look at these exponents, I want to get x^n everywhere, I see an x^n there, x^n there it is good and over here I had an x^n -2. 4044

I had to raise that up by 2 which meant raising all the n's in the formula by 2, that is what I did here and lowering that n by two.4057

Then I match the exponents, then I had to match the indices so I look at this I see the n=0, n=1 and n=0 here.4066

I want to pullout terms from lower 1 to get them to match higher ones, that is why I pulled out my n=0 term here and then I can start the series at n=1, the same thing here pullout the n=0 term so I can start the series at n=1.4079

That meant that now my indices are matched at n=1,that meant I could combine this series and this series and this series they are all combined together to give me this big series and it started at n=1.4094

They all had x^n so I factored out the terms from each part and there are couple of left over terms a0 and 2(a2) and so that is how it ended up there.4112

That three by the way came from that three right there and so the constant terms were equal zero that gave me an expression for a2 in terms of a0.4122

And then solving for the higher term in terms of the lower ones I got an + 2 in terms of an.4134

That is the recurrence relation that we are going to carry over into the next problem and use to actually figure out the coefficients of this series.4141

In example 6, we are going to use the recurrence relation that we derive in example 5 to actually solve the differential equation, remember we ended up with 2(a2) - 3(a0) equal zero then we have this recurrence relation.4154

Let me remind you that all of this is stuff we figured out back in example 5, if you took a break and you are just coming back you might want to go back and check example 5.4173

That will show you where these two formulas come from, they are not supposed to be a big mystery but we definitely are not working them out right now.4187

What we are going to do with these is go-ahead and solve higher coefficients in terms of lower ones, if we solve for this one up above ,if we solve for a2 to we get a2 is equal to 3(a0)/2.4196

We get a2 in terms of a0 and we get an + 2 in terms of an, and that was for n being bigger than or equal to 1, that again was coming from example 5.4215

Check back example 5 if that is looking mysterious to you, what I'm going to do is give myself a new page because it is going to take us a little bit of work to figure out the rest of this.4228

The recurrence relation is good for n being bigger than or equal to one, so lets see what coefficients we can figure out.4240

a0 there is nothing that is told me what a0 is, nothing in terms of lower coefficients and the same with a1 so a0 and a1 are arbitrary, I do not what they are.4249

I'm just going to leave them as a0 and a1 but then for a2, I figured out on the previous slide that a2 was equal to 3(a0) /2, that was from above, we worked that out, it actually came from example 5.4260

You can check back and see where that comes from and now the recurrence relation is going to kick in.4283

I'm going to start out by plugging in different values of n and starting at n=1 because that was the first value from which my recurrence relation is valid.4288

When I plug in n=1 to the recurrence relation, plug in n=1 right there I get a3=3(a1)/3 which simplifies down just 2(a1) n=2 gives me a4 =3(a2)/n+2, that is n=2, 3(a2) /4, but a2 was 3(a0)/2 so that is 3^2 x a0/2 x 4.4300

From n=3, I get my a5 is going to be 3(a3)/5 which in terms of a1 is 3(a1)/5, that is coming up from here when we figured out a3 in terms of a1, n=4 is going give me a6 is 3(a4)/6 which is 3^3(a0)/2 x 4 x 6.4348

n=5 is going to give me a7 is 3(a5)/7 which is in terms of a1 is 3^2 x a1/3 x 5, we are going to assemble this coefficients and build ourselves a couple series solutions.4393

Our original guess for the solution was the sum of (an)x^n, n starting at zero so this is a0 + a1x + a2x^2 + a3x^3 + a4x^4 and so on.4417

a0 there is nothing we can do about that because it was arbitrary, a1x nothing we can do with that, a2x(^2) no a2 was a3(0)/2, our a3 was a1x(^3), a4 was 3^2, a0/2 x 4, a5 was 3(a1).4439

There was an x^4 there plus 3a1/3 x 5 x^5 + 3^3 a0/2 x 4 x 6, x^ 6 + 3^2 a1, 3 x 5 x 7 x^ 7 and so on.4471

We can segregate out the a1 and a0 terms and the a1 terms, a0 it looks like it got 1 + 3/2 x^2 + 3^2/2 x 4 x^ 4 + 3^2 / 3 x 4 x 6 x^6 and a1 is x + x^2.4498

I'm going to write that as 3/3 to make a pattern work better later, plus x^5 the coefficient of x^5 is 3/3 x 5.4533

The coefficient of x^7 is 3^2 over 3× 5×7 and so on, that our series for a1 and we can write a nice pattern for this.4547

This this is a0 times the sum from n= 0 to infinity, it looks like I got these even numbers, remember that trick back from example 4, we go back and look at example 4, there is a trick to write this as 2^n x n factorial.4561

You can check that out in example 4, 1 over that and so what we have here is 3^n in the numerator, 2^n in the denominator, that is 3/2^n x x^2n.4580

I can actually write that as 3/2 x x^2^n/n factorial and then I do not have anything particularly good for my a1.4596

I'm just going write it in series form, the sum from n=1 to infinity, n=0 to infinity of 3^n/1 x 3 x 5 up to 2n + 1.4611

You could resolve this the same way we resolved it in example 4, I do not think it is really worth it that at this point.4628

What is worth it is to notice that this series on the left is exactly like the series for e^x remember that was x^n/n factorial and so what we have here on the left is a0 x e^3/2 x^2.4636

And there is nothing really good happening on the right.4658

I'm going to rewrite my two constants as c1 and c2, so c1 e^3/2 x^2 + c2 x this horrible series, n=0 to infinity of just 3^n/ the odd numbers, 2n + 1 x^ 2n + 1.4662

There is nothing really good that happens with that series on the right, you could fill in some even powers make a nice factorial but you still would not manage to convert into an elementary function.4692

That is our solution for that one, let me recap how we derive that, we started out with this recurrence relation that we had from example 5, there is no way you could have predicted that without having work through example 5.4705

We also had this expression about a2 in terms of a0 from example 5, that also came from previous work.4721

Now nothing here told us what a0 and a1 were, we had to leave those arbitrary, but then we figured out a2 in terms of a0 and then plug in a different values of n into this recurrence relation, n =1, 2, 3.4730

And gave us the higher coefficients in terms of the lower one, a3 came back to a1, a4 came back to a2 which come back to a0, a5 down to a3, down to a1, a6 down to a4, down to a0, and so on.4747

We got all this higher coefficients in terms of a0 and a1 so when we go back and we look at our original series or original guess, we can convert everything into a0 and a1.4761

We can factor out a0 times a bunch of terms, a1 times a bunch of terms and with a little clever accounting on these terms with a0.4775

Remember this was something that I covered in example 4 so you can go back and look and that if you do not remember how that worked.4787

This series on the left turned into e^x^2, 3/2x^2 series in the right did not really turned into anything good and we just had a leave it is a series.4794

We got these to our independent solutions and as usual I converted the a0 and a1 into c1 and c2 that is not really a necessary step you can leave it as a0 and a1, if you want.4806

That wraps up our lecture on series solutions for differential equations. My name is will Murray and you are watching www.educator.com, thanks for joining us.4820

Hi welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going study Euler equations, let us jump right in.0000

Euler equations have a very special form, they are the form x2 y″ + a constant which I’m calling α × XY′ + β (another constant) × y=0.0012

The idea there is that there is this pattern of descending powers on the x, this is very important.0027

We got an x2 here, we got an x here, and you want to think this is x 1, you want to think that this is x0 because it is just a constant.0036

You really have to have that exact pattern of the powers, x2, x and x0 in order to make the order of the equations work.0045

If you do have that pattern then there is a very easy solution it is basically xr.0054

Let me show you how to figure out what r is, what you are going to do is you are going to solve the characteristic equation for r and there is a little subtlety here.0060

Basically you take the coefficients from the order equation but there is a small change which is just instead taking α and β, you take α – 1 and β.0071

You solve this characteristic equation r2 + α – 1r + β=0.0083

That is just a quadratic polynomial, you can solve that by factoring, by quadratic formula, and whatever works for you to solve that and you will get 2 roots for r, you get an r1 and r2. 0090

This is not really hard, this is just algebra, you could get 2 real roots, so you could get r1=3, r2=5 or something like that.0106

You could get 1 repeated root, meaning you will get a double root or you could get 2 roots that are complex conjugates.0118

Something like a + π and a – π, that is if you are on a quadratic formula and you get a negative number under the square root sign, then you are going to get 2 complex conjugate roots.0125

Let me tell you what to do in each one of those situations and how you will write down the solution to the differential equation.0136

If you have real distinct roots, meaning 2 different real roots the general solution to the equation is just xr1 × constant and xr2 × another constant.0144

It is very easy there if you have repeated roots, same copy of root appears you have xr1 and xr1 × natural log of x.0155

Again, multiplying each one by a constant and if you have complex roots then it will always occur in conjugate pairs.0166

Remember that the quadratic formula is something (+ or -) the square root of something.0173

If you get a negative number under a square root then you will end up getting a conjugate pair, something like a (+ or -) π.0182

It is a little more complicated if you do that but it is just a formula to remember.0191

It is xa × cos(x) of logx to the b and xa × sin of natural log of xb.0195

Those are your 2 fundamental solutions and you just multiply each one by an arbitrary constant and you will get your general solution.0208

That is pretty much all you need to know for Euler equations, you solve the characteristic equation, get the 2 roots.0215

Just depending on which of these 3 situations it is, you just drop it in to one of these 3 formulas, let us try that out with some examples.0223

In our first example, I want to find the general solution to x2 y″ – 3xy′ + 3y=0.0231

We got this Euler pattern on the coefficient x2, x and then a constant here, so it is an Euler equation.0240

The α here is -3 because that is the coefficient here, the β is 3 and remember the characteristic equation is always r2 + α- 1r + β=0.0249

That is the equation that you always solve for r and in this case we have r2,0272

Now α is -3, so α – 1 is -4, this is actually -4r and β is 3, +3 =0.0280

That equation factors quite nicely that is r-1 × r-3=0.0292

Our 2 values of r are 1 and 3, r=1 and r=3, now we can drop those right into the general solution.0303

The general solution is c1 × x + c2 x3, that is the solution to the differential equation.0317

It is a really nice relief that this Euler’s equations are much easier than some of the other things that we have been doing recently like series solution.0336

Let me recap quickly how we worked that out, we noticed that this was an Euler equation because it has that descending pattern of powers x2, x, and constant.0345

We read off our values of α and β, just the coefficients there -3 and 3.0356

Drop them into this characteristic equation, the key point there is to remember to subtract 1 from the alphas.0362

We got r2 -4 +3 =0, factor it out, get our roots and those become the powers and we get our general solution.0367

Let us go ahead and try another one, we still have an Euler equation here because the powers on x is descending.0380

We have x2, x, and a constant, our α here is -7, β is 16.0389

We are going to use our generic characteristic equation r2 + α -1r + β=0.0402

Our α is -7 so α – 1 is -8, -8r + 16 = 0.0412

(r) factors into r-42=0, we get the double root at r =4 or 4.0422

We had a different formula for double root.0433

The general solution c1 x4 but we can not have c2 x4 because that would just be a copy of our first solution.0439

C2 x4 natural log of x, we got that straight from the lesson overview at the beginning.0449

Whenever you have that double root, that second solution you just multiply it up by natural log of x.0459

That is the end of that one but just to recap we noticed our descending powers of x.0467

X2, x, and a constant, that is what triggers you that it is an Euler equation.0473

Write down your α is the coefficient of first, β is the coefficient of the second one.0478

Plot those into our generic characteristic equation where you have an α -1 so that -7 turns into -8, factor it out.0483

Once you get a double root for r you know one of your solutions is x4, but the other solution, in order to get an independent solution you got to multiply it by natural log of x.0493

That is our second independent solution there.0503

Our next example here, again we have Euler equation, we can check that by noticing the descending powers of x, x2, x, and a constant.0508

We got the descending powers of x, we got an Euler equation.0519

α here is -1, β is 5, those are the values we are going to use in the characteristic equation.0522

R2 + α – 1r + β= 0, if α is -1, then α -1 is 2, β is still 5.0533

Now that thing does not factor so we are going to jump in the quadratic formula to solve that.0551

Remember the quadratic formula says –b (+ or -) square root of b2 - 4ac/2a.0557

Here our b is -2 so our- b is +2 (+ or -) b2 is -22 so that is 4.0573

4ac is 4 × 1 × 5=20, over 2a is 2, that is 2 (+ or -) 4 – 20 is -16.0584

The square root of -16 is 4i, I’m going to get complex roots here, that simplifies down 1 (+ or-) 2i.0599

Remember we have a generic solution formula when you have complex roots.0614

The generic solution formula, let me remind you what that was from the lesson overview, the beginning of class.0620

C1 xa × cos natural log xb + c2 xa × sin of natural log of xb.0627

You could also use your natural log rules and write this as b natural log x inside, that will be ok if you look that better.0648

The a and b here, they are not the a and b from the original equation so be careful about that.0655

The a and b come from the a (+ or -) π here, it is not the same as these a’s and b’s that we used in the quadratic formula, be very careful about that.0661

What we have here is our general solution, c1 xa, our a is 1, I just write x × cos of natural log of xb, b is 2 here,0676

So natural log of x2 + c2 x × sin of natural log of x2 and that is our solution.0690

Let us take a look and see how that worked out, we had our Euler equation because of those descending powers of x, x2, x, and a constant.0712

We read off our coefficients, α is -1 the coefficient of the x term, β is 5, we plug those into the characteristic equation.0721

Remember the subtlety where the characteristic equation you drop the α down by 1.0732

α -1 becomes -2 and β is 5, to solve that we have to go to our quadratic formula.0737

That simplifies down to 1 (+ or-) 2i and we use that as our a and b, not the a and b from the quadratic formula but the 1 and the 2 are a and b,0747

That we are going to plug into our generic solution for Euler equation when you have complex roots.0760

We plug in x1 here and we get (x cos log of x2) and (x sin log of x2) as our 2 independent solutions.0768

We put them together with constants to make our general solution.0779

For example 4, we got x2 y″ – 6xy′ – 12 y=00785

We check out first that this satisfies the properties of being Euler equation, those descending powers on x, (x2, x, and a constant), (x2, x1, x0).0793

Remember it got to fit that format exactly to be an Euler equation, if it does not fit that format then you are really out of luck, you got to use some other techniques which certainly be more complicated.0805

It is worth checking if something is an Euler equation if so, these things are real pretty quick to solve.0817

If not then you got to have to do something that is probably be much more lengthy.0823

Since this one does work, we read off our α as -6, our β is 12, and we set up our characteristic equation.0827

R2 + α – 1, r + β= 0, that is our squared.0837

α -1 is -7, r + 12=0, that is another easy one to factor, that is r-3 × r-4=0.0846

R is 3 or 4, so 2 distinct real roots that is the easiest situation for an Euler equation.0861

Right away I can write down my general solution it is just c1 x3 form that first root + c2 x4 from that second root.0871

Just like that we are done with that one.0884

Let me remind you of how that one worked out, we first check the powers x2, x1, x0.0893

That looks like an Euler equation so that means I can write down the coefficients.0900

α=-6, β=12, drop them right into my characteristic equation which has that shift to the α down by 1.0904

We get r2 - 7r that is α-1 + 12=0, factors nicely and gives me couple of roots.0914

Those roots become the exponents on x for the general solution.0923

On example 5 here, again we have an Euler equation because we recognized this descending power of x’s.0932

X2, x1,and x0 here, no x at all means x0 or you think of it as being a constant.0939

I can read off my coefficients, α is 5, β is 4, I’m going to set up my characteristic equation.0948

It always has a form r2 + α -1r + β=0.0958

Remember to shift that α by 1, that is the subtlety of Euler equations that we have not really seen anywhere else.0967

In this case the α is 5, we get r2 + 4 r + 4 =0, that first 4 here, this one right here, that comes from 5-1, it has nothing to do with β being 4 there.0973

That is easy to solve, that is r + 22=0, r=-2 and that is a double root.0988

Remember we have a special formula for one you have a double root of an Euler equation and that formula is, first you have x to that power, c1x to that power.1001

But then you can not have c2 × x to that power again because that would just be a copy of your first solution.1013

The way you get around that is you multiply on a natural log of x.1021

That is our general solution to that one.1028

Let us recap how that problem worked out, we recognized the descending powers x2, x1, x0, ok that means it is an Euler equation.1039

We identify our α as a coefficient of the x term, β is our coefficient of the constant term.1051

Then we go to our generic characteristic equation with an α -1 in it and we get r2 + 4r + 4, that is using α -1 to get that 4 right there.1059

That factors easily into r + 22 = 0 that turns into r=-2, that is a double root.1075

In case of a double root, our solution is just like with 2 distinct roots, we form x to that power, x-2.1086

For the second one, we can not just form x-2 because that would be a copy of our first solution.1096

We go x-2 natural log of x to get ourselves a distinct solution and we put a constant on each one.1101

On example 6, we got the differential equation x2 y″ – 3xy′ + 29y=0.1112

As before, look at this we got x2, x1, x0, that means it is an Euler equation so we can use what we have been learning in this lecture.1122

We are going to identify our α and our β, α is -3 here, make sure you include the negative, sometimes people just see the 3 here and say ok α is 3.1135

Of course that is going to lead you to disaster later on, do not forget those negatives if they are part of it.1145

β is 29, we go to our generic characteristic equation r2 + α – 1 r + β =0.1152

α is -3, α -1 is -4, r + 29 =0 and it will be great if that would factor but it does not factor easily so we are going to the quadratic formula for that.1166

R=-b (+or -) the square root of b2 - 4ac / 2a.1185

In this case our b is -4 so –b is +4, (+ or -), b2 is 16 and 4ac is 4 × 1 × 29 which is 116, that is all divided by 2a but a is just 1 so that is 2.1196

That is 4 (+ or -) the square root of -100/2, now -100 the square root of that is 10i.1224

This is 4 (+ or -) 10i/2, 2 (+ or -) 5i and we have a generic solution formula for when we get complex roots to the characteristic equation.1237

It is c1 xa × cos of natural log of xb + c2 xa × sin of natural log of xb.1256

You could also write that as b natural log of x, that would be correct.1273

The a and b, they are not the a and b from back here in the quadratic formula, they are the a and b that you got in the end.1279

Let me go ahead and plug those in, the general solution here is (c1 × x2),( cos r of natural log5 + c2 x 2 × sin of natural log of x5.1290

You could write that as 5 natural log of x if you like.1312

That is our general solution to this Euler equation.1319

Let us go back and see how that worked out, first thing to do is recognize the powers, x2, x1,and x0.1329

That is what you have to have for an Euler equation to work out as those descending powers of x’s.1337

If you do its going to work out easily, if you do not you have to abandon the Euler methods completely and you have to try something completely different.1343

We found that it worked out for this example, we identify our α is the coefficient of the x term, our β is the coefficient of the constant term.1355

We go back to the generic characteristic equation r2 + α – 1r + β = 0.1372

Our α is -3 so subtracting 1 from that is where this -4 came from, we got this quadratic equation that really did not factor nicely.1381

So we use the quadratic formula, we got a –number under the square root so we get complex numbers 2 (+ or -) 5i.1391

Then we drop those as our a and b, not the a and b from the quadratic formula but our a and b from the complex number we found.1400

We drop those in to our generic solution formula for Euler equation for complex roots, which is xa cos log b + xa sin log of xb.1409

We get (c1 x2 cos log of x5) (c2 x2 sin of log of x5).1423

That is our general solution to the differential equation.1431

That is the end of our lecture on Euler equations, by the way you might have come here looking for Euler methods which is actually something quite different.1436

That is something we are going to get in to a later lecture here on the differential equations lectures.1446

If you are looking for Euler methods here and this did not look at all like what you are expecting, just scroll through and look for a later lecture on Euler methods.1451

That is a numerical solution to differential equation and you will find that here on www.educator.com.1459

It is just that this was a specific lecture about Euler equations.1465

That wraps everything up here, this has been the Euler equations lecture in the differential equations series here on www.educator.com.1470

My name is Will Murray and I thank you very much for watching, bye.1479

Hi and welcome back to www.educator.com these are the differential equations lectures and my name is Will Murray.0000

Today we are going to study series solutions around regular singular points.0006

First, we have to learn what all those terms mean, the idea is that we are going to try to find series solutions to the differential equation P of XY″ plus q(XY)′ + r(xy) = 0.0011

The first thing you want to do there is to divide away this coefficient in front of Y″, we are just going to divide that into the denominators of the other two terms.0027

We get Y″ + q(x)/p(x)Y′ + r(x)/p(xy)=0, now we would like to study series solutions around x0=zero.0038

This series that we have been studying is really a short hand for x - x0^n but then we have been plugging it x0 = 0,.0052

We want to study the series solutions around x0 is equal to 0 but a problem arises if when we plug in that value of x we get a 0 in these denominators here.0061

What we are really worried about is the case when p(0)=0 and when that happens we call X0=0 a singular point of the differential equation.0074

Our original series strategy is the one we learned a couple of lectures ago do not work anymore, that is when we have to modify our strategy.0090

That is really what this lecture is all about, our original series strategy is something we covered in an earlier lecture but now we are talking specifically about series solutions around singular points.0099

That is where we have to learn some terminology before we can learn anymore, the main definition we have to learn is what a pull of order n means.0112

The idea there is it really a way of measuring what happens essentially when you have a 0 in the denominator and measuring how badly you have a 0 in the denominator.0125

In particular, what we do is we say f(x) has a pull of order n, if we write a series version of f, if the first term is one over x^n, let us see what we mean by that.0138

We will do some examples later on with more details, we may not give a very quick example here if we had a function whose series expanded out into a 1/x^2 + 2 /x + a constant + 3x.0154

And then it continued into positive powers of the series, I would look at that note and say okay the first term there is 1/x^2 .0173

The worst explosion, the worst case of dividing by zero has x^2 in the denominator that would be a pull of order 2.0184

Let me remind you what kind of differential equations we are trying to solve, we are trying to find series solutions to Y″ plus remember we had a q/ (p)Y′ + r/py=0. 0199

That was the differential equation that we are trying to solve from the previous slide and the point is that we are trying to measure just how badly we are trying to divide by 0 when we have this P in the denominator.0220

In particular, what we do is we look at this the first bad term there the q/p and we would like that to have a pull of order at most 1.0233

The division by 0, the order of the division by 0 is it most 1, the other term there is r/p and we would like that to have a pull of order at most 2.0246

The division by 0 there is allowed to go up to x^2 in the denominator but nothing worse than.0263

If those two conditions hold then we say x0=0 is a regular singular point of the differential equation.0270

That is our definition of regular singular point, we are going to be checking the orders of the pulls at those two functions the q/p and the r/p.0280

It turns out that if we have a regular singular point that we can still use a series strategy to solve our differential equation, what we do is we take our original series which was a an/x^n.0290

If you expanded out that looks like a0 + a1x + 2x^2 and so on and We bump it up by some extra powers of x. 0304

We multiply it by x^r and we don't know what value r is yet but we will figure it out in the course of solving the differential equation 0314

Our strategy is essentially the same as before except that we have this extra factor of x^r in our series.0326

A lot of times we will bring this x^r right into the series and combine it with x^n , I have something like n=0 to infinity of a(sub)/x^n + r.0335

That is going to be our new guess for y that we are going to use for these regular singular points and then we are going to plug that in and we are going to try to solve it.0349

I will give you more details in a moment but there is one more issue that I want to cover right now which is that we make a little assumption here.0356

Notice that if this a0 was 0, if that term happened to be 0, then what we could do is we could factor at least one more power of x outside and combine that with x^r0364

In order to prevent that from happening since we are already going to figure out what x^r is, we are going to make a little assumption here which is to assume that a0 is non zero.0378

Because if a0 were 0, we could just factor out more powers of x, we are essentially saying that the case where a0 is 0 can not happen.0395

We are going to assume that a0 is non zero and that can help us a little bit in our arithmetic later on.0405

The take away from this is that we are going to make this assumption y= a/n x n x with a0 non zero.0413

Let me go ahead and show you what we are going to do with that.0425

We are going to plug that series into the differential equation and there is going to be a lot of algebra that goes on while we try to reconcile the different terms of the different series.0429

And try to combine them all into a single large series., that is pretty similar to the series solutions that we already studied in the earlier lecture.0440

At the end we will get what is called an indicial equation for r.0449

Let me highlight this, we will get an indicial equation for r, actually it will just be a polynomial equation, it will be a quadratic equation, something like r^2 + 3r +2=0.0455

It will be something relatively easy to solve like r^2 + 3r +2 =0, relatively easy to solve and we will find two different roots for r.0474

What we do with those two roots is we check the difference between them, we subtract those two roots from each other.0486

It turns out that if the difference between those two roots is an integer then we can only find a solution for one of the roots, we can find a solution for the larger roots only.0493

We will see an example of this, we will work out some solutions but if the difference between the two roots is not an integer, if it is a fraction or something else then we can find a solution for each one of the two roots.0506

We will see how that works out as we go ahead and try some examples, let us try that out in the first example, we are just practicing understanding this definition of the order of a pull.0520

We got a bunch of different functions here and for each one we want to classify the order of the pull.0536

Let us go ahead and work that out, the first one is e^x/x and remember that e^x, I know series for e^x, it is 1 + x + x^2/2 factorial + x^3/3 factorial and so on.0541

That is something that we essentially memorized when we are learning about Taylor series all the way back in calculus 2 and then we reviewed it earlier on in the differential equations lectures.0562

The first lecture about series was just a review of Taylor series to remind you that we are to be using things like this e^x/x.0572

That is multiplying everything here by 1/x so it is 1/x x 1+ x + x^2/2 factorial and so on and if I distribute that 1/x, I get 1/x +1+ x/2 factorial and so on. 0584

Now I see that I got a series for my function and the very first term in the series is 1/x^1, so that means I have a pull of order, order 1.0604

That is my answer for the first function there is that I have a pull of order1, we will stick to the next one sin(x)/x.0624

I remember that I have a prefabricated Taylor series for sin(x) it is x - x^3/3 factorial + x^5/5 factorial and it goes on like that with odd powers of x.0632

sin(x)/x is 1/x x that series x - x^3/3 factorial + x^5/5 factorial and it keeps going there.0649

I'm going to distribute that 1/x and I will just get 1 - x^2/3 factorial + x^4/5 factorial and so on.0669

If I look at that series there are no terms with x in the denominator, the very first term there is a 1 and we can think of that as x^0 ,that means what we have here is a pull of order 0 for that function.0684

Our next function is very easy, it is 1/x^2 and you can really think of that is being it already and series form, you can write that is 1/x^2 +0/x + 0 + 0x and so on.0705

We see that the first term with x in the denominator is that 1/x^2, so the exponent on the x there is 2.0723

We say here that we have a pull of order 2 and for our last function the 5x + 4/x^3 -2/x 5x + 4/x^3 - 2/x.0739

I'm just going to write that in ascending powers of x, the smallest one x^-3, 4/x^3 - 2/x + 5x, I have written it in ascending powers of x.0760

The powers x are going from -3, -1, +1 and I see that the first one there is an x^ -3.0775

That is where this series starts so that is a pull of order 3, so let us recap what we did there.0784

In each case, we tried to write the function that we were given as a series and then we looked at the power of x in the denominator in the very first term of each series.0799

In the first case, we had e^x/x, I know this series for e^x, we learned that back in calculus 2 and I reviewed it at the beginning of this series chapter of differential equations.0811

And then I multiply that by 1/x, distribute it to 1/x and I got series where 1/x is the first term.0825

That is x^ -1, so my first negative power of x is -1 is 1, my pull there has order 1.0832

For sin(x), I remembered a Taylor series from calculus 2, we reviewed that back in the first lecture series here in the differential equations. 0844

Multiply that by 1/x, distribute the 1/x and I get a series that starts at 1 which is the same as x^0, so I say that is a pull of order 0.0853

1/x^2 is a tiny little series by itself, you think of all the other terms as being 0, the first term that we see there is 1/x^2 power and we have a pull of order 2.0865

Finally, this 5x + 4/x^3 - 2/x we just arrange that in increasing powers and we see that the first term there will be 4/x^3.0879

It is that 3 right there that we are focusing on that, tells us that we have a pull of order 3.0890

In example 2, we have to determine whether x0=0 is a regular singular point for each of the following differential equations.0898

Let me remind you quickly what were looking, when we look at regular singular point we write our differential equation and we always divide by that leading term of r or by p.0906

Sorry so we get Y′ + q/p Y″ + q/pY′ +r/p, y=0.0920

That is the format that we have to get these differential equations into and then what we do is we look at q/p and r/p and in order to be a regular singular point.0933

The q/p has to be a pull of order at most 1, the r/p has to have a pull of order at most 2 and 0, those are the conditions were looking for.0944

Let us check out the differential equations we have been given here and see if it satisfies those conditions.0966

On the first differential equation we got x^2 Y″, right away I'm going to divide it by the x^2.0973

I will label this as a and b here. 0979

The first one a here, I'm going to divide by the x^2 and I get Y″ + sin(x)/x^2 Y′ + 3/x^2, y=0.0982

I need to look at those two coefficient functions the sin(x)/x^2, we will start with the sin(x)/x^2, sin(x) I remember my series is x-x^3/3 factorial + x^5/factorial.0999

I memorize that back in calculus 2 and we reviewed that back in the beginning of differential equations.1020

sin(x)^2 that is 1/x^2 sin(x) is multiplying 1/x^2 by each term one of 1/x^2 x x is 1/x - 1/x^2 x x^3 is just x/3 factorial + x^3/5 factorial.1028

I have lowered the power of x by 2 in each case and what I see here is a series that starts at 1/x that is like 1/x^1, I see a pull of order 1.1051

I see that that is ok to be a regular singular point but I got to check the other one as well so the 3/x^2 that is the r/p, 3/x^2.1069

That is already a series by itself it is just 3/x^2 + 0/x + a constant term is 0 + 0x and so on.1082

The 3/x^2 is the first term there that tells me that I have a pull of order 2, my q/p had a pull of order 1 and my r/p had a pull of order 2.1094

That is allowed for the definition of regular singular points, what that tells me is that this is a regular singular point.1115

My answer there is that x0=0 is a regular singular point of that first equation.1141

For the second equation, let us work out equation b here, we divide by x^2 and we go Y″ + cos(x)/x^2 Y′ + e^x/x^2, y=0.1148

Let us look at q/p and r/p and see what the orders of the pulls r for each of those functions.1171

cos(x) I have a series memorize back from calculus 2 and we reviewed it in the first differential equations lecture on series.1178

That is 1 - x^2/2 factorial + x^ 4/4 factorial and so on and cos(x)/x^2, remember this is q/p that is why we are looking at that because we want to check the order of the pull there.1189

If we divide each of those terms above by x^2 we get 1/x^2 -1/2 factorial + x^2/4 factorial and so on.1211

This is already in series form, I look at it and see the first term right there is my tip off that I have a pull of order 2.1224

Let us go ahead and check the r/p that is coming from e^x/x^2 , e^x is 1 + x + x^2/2 factorial + x^3/3 factorial and so on.1239

e^x/x^2, the reason I'm looking at that is because that is my r/p, I need to check the order of the pull there to figure out if it is a regular singular point.1258

I'm going to divide each term by x^2 above so I get 1/x^2 + x/x^2 is 1/x + 1/2 factorial + x/3 factorial and so on.1274

I look at that, I see the first term there is 1/x^2 so that tells me that I have a pull of order 2, so both q/p and r/p have pulls in order 2 for this second equation .1292

If I check my definition here, my definitions as I need a pull of order at most 1 for the q/p and at most 2 for the r/p.1310

I forgot to write that was a pull of order 2, so the r/p having a poll of order 2 that is totally fine there.1328

But the q/p having a pull of order 2, that violates my definition because it was supposed to be a pull of order at most 1.1337

That tells me that x0=0 is not a regular singular point for this differential equation.1344

The reason we are studying this, is because we have a solution strategy that works for regular singular points and it does not work for points that are not regular singular. 1364

Our solution strategy would work for this first differential equation, it would not work for the second differential equation , that is why we are studying these now. 1375

That is all we had to do for this example but let us go back and recap, for each one, the first thing we did was divide away this x^2 into the denominator.1385

We have a nice clean coefficient on the Y″, that is what we did here, we divided by x^2 and then we looked at q/p and r/p.1396

We wrote series for each one, I remember my series for sin(x) so I divide that by x^2 to get a series for q/p.1406

Here is my series, I see the first term has 1/x^1, it is a pull of order 1 and that is allowed in my definition.1414

The r/p is 3/x^2 which is a series by itself, it starts with x^2 in the denominator so it is a pull of order 2, that is also allowed in the definition which is why I say it is a regular singular point.1424

For the second one, again I divide by x^2 which gives me cos(x)/x^2 for q/p and e^x/x^2 for r/p.1439

I try to write a series for each one here is my series for cos(x) when I divide by x^2 that turns a 1 into 1/x^2 and it drops everything else down by 2 powers of x.1449

The first term I see there is cos(x)/x^2, so I have a pull of order 2, now that is not allowed by the definition so already I would know that that is not a regular singular point.1461

But I went ahead and check the other one, just for pedagogical reasons I guess. e^x we know we have a Maclaurin series for e^x here.1473

That is something I remembered from calculus 2 and then e^x/x^2 , I just drop everything down by 2 powers of x so it starts out at 1/x^2 which is a pull of 2.1485

Actually the r/p is okay, it fits the definition but it was the q/p that mess things up there because it was a pull of order 2 and was only allowed to be a pull of order 1.1496

That is why that second differential equation x0 =0 is not a regular singular point, it is really because the q/p ruin things there. 1509

Hopefully now you are getting a little more comfortable with calculating the orders of pulls and classifying whether something is a regular singular point or not a regular singular point.1520

The next step here is to actually use these series techniques to solve some differential equations so let us practice that with the next example.1530

Here we are going to find and solve the indicial equation for this fairly complicated differential equation.1543

Let me note first that we actually do have a regular singular point because we have the q/p, that would be if you divide 3x by 2x^2 which would give you 3/2x.1551

The series there if you think of that as being a series, it has just has one term and that term has x above one in the denominator so that is a pull of order 1.1572

The r/p is 2x^2 -1 / 2x^2 which you can distribute and break up those terms into 1-1/2x^2, so that the first term there, we would order things by ascending powers of x.1588

That is a pull of order 2 and so collectively those mean that we had satisfied the conditions, remember q/p is allowed to have a pull of order .1612

r/p is allowed to have a pull of order 2, we have a regular singular point which means we can use our series strategy to solve this differential equation. 1627

Let us remember what that series strategy is, we said we take our normal old-fashioned guess for series which was n=0 to infinity of (an)/x^n and then we multiply the thing by x^r.1644

I'm going to jack that up by the power of r, that is the new element there, instead of having x^n we have x^n + r.1661

That is our new strategy for using series to solve differential equations around regular singular points and let me go ahead and look at what I'm going to have to do when I plugged this y/n.1670

I see I'm going to have a -y there so I'm going to go ahead and write -y is just the same thing with a negative sign.1685

n=0 to infinity of -a/n(x^n + r), I also see them and I'm going to have a 2x^2 x y, let me go ahead and calculate that.1694

2x^2 x y is the sum from n=0 of 2(an) and then y was x^n + r, that x^2 is going to bump it up by 2 more powers.1707

We will have x^n + r + 2 when we work out 2x^2 x y, let me go ahead start writing some derivatives down.1724

Y′, if I look back at my original y is the sum from n=0 of n + r, that is my exponent times (an) x x^n + r -1.1734

We mentioned a difference at this stage from our earlier series solutions, our earlier series solution did not have the extra term of r in the exponent.1753

What that meant was at this stage we had just n instead of n + r in front of the (an).1766

What we noticed in the earlier solutions was that because the first term is n=0 we could drop out that first term and start at n=1.1774

We can not do that anymore because this r messes up that strategy so we can no longer-1784

Let me put a red warning here, we can no longer change that n=0 to n=1 right away because we can no longer drop out the n=0 term, it is no longer equal to 0.1792

That is a little warning there, let us go ahead and look ahead at what are we going to be doing with this Y′, we are going to be calculating 3x x Y′.1808

Let me figure that out 3x x Y′, I should have said Y′ here this was the derivative, 3x x′ is the sum from n=0 of 3 x n + r x (an).1818

We have multiplied it by x, instead of x^n + r - 1, I got one more power of x so it goes back up to x^n + r and now let us figure out Y″.1838

Y″ is one more derivative, again we have to start it at n=0 in the previous series solutions we would have started this at n=2 because the first couple of terms dropout automatically.1854

That is no longer true, so we are going to have n + r another power comes out, we have n + r -1 times (an) x^n + r -2.1865

That is our Y″ and when we will look in the differential equation I see Y″ is going to get multiplied by 2x^2.1882

Let me go ahead and figure that out, 2x^2 Y″ is the sum from n=0 of n + r.1890

Everything is the same here, n + r -1(an) and now I had x^n + r - 2 before, but I have multiply that by x^2 so I get x^n + r and I see I also need to put my two in there.1900

Let me squeeze that in here because we are multiplying by 2x^2 so we have a n extra factor of t2 in there.1920

These are the series we are going to have to resolve and combine in the differential equation, we have a -y, we have a 2x^2y, we have a 3xy′ and we have a 2x^2Y″.1931

We need to resolve all of these series with each other and what I have to do here is to get the exponents to match each other and get the starting coefficients to match each other.1953

This is going back to the original strategy on series which we covered in another lecture. 1969

If you are a little rusty on that, maybe go back and watch the other lecture on series solutions because I'm following the same strategy here.1975

It was a 2 step strategy, step one was to match exponents on x and then step two was to match the starting indices.1983

I'm going to be using that here and if you are rusty on the general strategy there, we do have another whole lecture on series solutions on differential equations.2004

I think it was 2 lectures ago, maybe go back and check that out and hopefully this will make a little better sense to you.2018

In the meantime, I got these 4 series this one has an x^n + r, this one has x^n + r + 2, this is x^n + r and this x^n + r.2025

I'm going to the try to fix this one x^n + r + 2 and the way I'm going to do that is using my little trick that we learned back in the series lecture where I raise the index by two.2038

I'm going to raise that up to n=2 and I lower all the n's in the formula by 2, I have 2, a (sub)n -2 and n^x + r + 2 drop that down by 2 and I just get x^n + r.2052

That is really nice because, sorry n + r not n + 2 there, that is really nice because it means now all my exponents are x^n + r.2071

I have achieved the first step of my strategy there, second step is to match the starting indices, what I'm going to do is look at the starting indices I see here, I got n=0, n=2, n=0.2084

What I'm going to have to do is pull out starting terms from all the lower ones in order to match up the highest one.2103

The highest one is this n=2 , I'm going to go back to the other ones and pull out some initial terms in order so that I can start all the series at n=2.2110

Let us look at this one, if I pull out the n=0 term I get -a(sub)0, x^r, that is my n=0 term.2122

My n=1 term is -a(sub)1 x^r + 1 and then I can keep going with a series starting at n=2 then I had -(an)x^n + r.2137

The next one, I already started at n=2 , that is ok that one starts at n=0. I got to pull out a couple of terms here.2154

I pull out the n=0 term that is 3(ra)0 x^n and are sorry x^r, the n=1 term is 3 x r + 1, a1x^r + 1, I will plug in x^1 there.2162

I then still have the rest of the series which is now starting at n=2 of 3 x n + r(an) x^n + r.2188

Finally, this last series also starts at n=0 so let me pull out 2 terms here when I plug-in n=0, I get 2 x r x r -1 plug-in n=0 x a0(x^r), if I plug-in and x=, I get 2 x r + 1 x just r x a1 x x^r + 1.2203

I get the rest of the series plus the sum from n=0 of 2 x n + r - 1 an(x^n + r) , this is really nice what I see that I have here is all of my remaining series.2244

I have that one right there and let us see, I have this one and this one and this one I will admit a little mistake here I should start that it in n=2 because what I had just on was pulled out, The n=0 and n=1 term so let me recap here's a series now started at n=22267

We got x^n = r this series has n=2x an x^n + r, this series has (an) = to a x^n +r and this series has an = (an) x^n + r, all my series started at n=2 and they had the exponent of x^n + r.2290

I matched my starting indices here at n=2, I match my exponents, I have matched my starting indices and I'm ready to combine my series.2311

What I'm going to do is combine these beginning terms that I have on the outside, let me just look at those beginning terms. 2326

I see I'm going to have several terms with x^r and a0 and what do I have multiplied by that, up here I have a negative 1, here I have a 3r + 3r.2339

Let us see here I have 2r x r - 1 + 2r x r - 1, that is all times a0 x x^r, let us see, I'm going to have some terms of A1 x x^r +1 so let me group all those together.2359

I have got a minus A1 here so -1 I got a 3r + 1 + 3 x r + 1 and here I got a 2 x r +1 x r so +2 x r +1 x r that is all x A1 x x^ r + 1.2386

Now I'm going to combine all the series in the boxes the nice thing about them all is that they all started at n=2, I very carefully arrange that and they are all I can have a term of x^n + r.2417

Let us see what kind of coefficient I can get out of each one here, I see have a -(an) here actually I'm going to have (an) on several of these terms.2434

Maybe I will go ahead and factor that, I have an (an) on several of these terms, I will have -1 x an that is from this one here, this one has a 2(an)-2, I will write that out separately plus 2(an) - 2.2446

Here is another term with (an) , 3n + r +3 x (n+r) and now here's another term with (an) +2 x n + r x n + r - 1. 2466

That is what I get when I plug all of these series into my differential equation and let us carry on with these terms, what I'm going to find in the next slide is that I am assuming that a0 is non zero.2491

Remember we made that assumption way back earlier in the lecture, we are going to set this term equal to zero, and we will see where that takes us that is going to be the indicial equation.2506

Let me remind you what that equation we just had was, since a0 is nonzero that term that I just underlined on the previous slide was -1+ 3R +2 x r x r -1 =0. 2517

This right here is the indicial equation. 2541

That is the indicial equation and we are going to use that to solve for our values of r, let us go ahead and simplify that, it is actually a pretty easy quadratic polynomial.2554

We get -1+ 3R + 2r^2 - 2r=0 so I got 2r^2 + 3r - 2r + r -1 =0. 2564

You could use the quadratic formula to solve this, this one factors pretty nicely, I'm just going to go ahead and factor it to solve it, I see that I can do 2r -1 x r + 1 will work because that would have a -r and a +2r.2581

That would give me a +r in the middle.2602

That is a way to factor that and if I set each of these equal to zero I get 2r -1=0 so r is 1/2 or r + 1=0.2604

r= -1 and so those are my two values for the roots of the indicial equation and so that is actually all we were asked to do for this problem.2621

Let me just recap what we did there, it was kind of complicated but we started out with this same guess that we are going to use every time.2635

y= the sum from n=0 to infinity of (an) and here is the difference from the previous series solutions x^n + r, we wrote that I guess for y we figure out Y′ we figure out Y″.2643

We got some big nasty series for each one than for all the series we got, we plug them into the differential equation which means multiplying them by various powers of x Y″ by 2x^2.2661

We figure out 2x^2 Y″, we figured out 3x Y′, we figured out 2x^2 x y.2676

We got all these different series and the first step with them was to match the exponents so that all of those series had the form while they all had different coefficients x^n + r.2688

They all had that form so we match the exponents at n + r and then the second step was to match the starting indices on each one.2707

The problem was that we had n=0 on some of them and n=2 on some of them, what we did was we took the n=0 term series and we pulled the n=0 term and the n = 1 term.2725

We just wrote those separately and then we could start our series at n=2 so once we got both the exponents and the starting indices to match then we plug them into the differential equation.2742

We plugged all the series into the differential equation and we got several terms with a0 x x^r, we got several terms with a1 x x^r + 1.2762

We got a big term n=2 x x^ n + r and we knew that a0 was non zero, that was our assumption back at the beginning of the lecture and so all this coefficient of a0 x^r that had to be equal to 0.2780

That is where we got this equation right here which was the indicial equation.2805

That is how we got that, once we got that it was a relatively easy quadratic equation we solve that out and get a couple values of r and that is actually what we had to do for this problem.2811

In the next example, we are going to keep going the same differential equation, keep going with this same long series solution that we've been working so hard to obtain.2822

We will actually go through and pick a value of r and we will solve it out and will get an actual solution.2832

Let us see how that works in the next example, in example 4, we are going to find the solution to the differential equation above corresponding to one of the two roots of the indicial equation.2840

Let me remind you this is the same differential equation that we use for example 3, we did a lot of work to write down a series solution we started out with y= the sum of (an)x^n + r.2851

We did a lot of work plugging that in and we figure out that r could be 1/2 or r it could be -1, in this example we are going to pick up where that example left off.2867

All of this is exactly what we got when we plugged our series solutions into the differential equation.2880

If this is all looking like gibberish, go back in and check out example 3 again because all this is coming from example 3. 2890

You will see how we derived all of this in example 3 and example 4, we are going to keep going forward and working with it and trying to actually get to a solution.2901

With our indicial equation here, we figured out that r= -1 so we are going to plug in r= -1 back into these various terms.2917

We already figured out that if r= -1 we plug that back in here we get -1+3 times -1+2 x -1, now r -1 is -2 that times a0 is equal to zero.2929

If you simplify these terms here, we get -1-3+ 4, you just get zero a non-equals zero which tells you nothing. 2951

We get no information about a0, so a0 is arbitrary.2968

We get no information about a0, a0 is going to have to just stand by itself for a while but we are also been plugging in those r -1 into that first set there.2978

We are also going to plug in r= -1 into this second term, so we get -1+3 x r + 1 would just be zero and then +2 x r + 1 again would be zero x -1 that times a1=0 is.2989

That term drops out, we does get -a1=0 so our a1=0, we sort of analyze as much as we can about the first two coefficients of the series, we got no information about a0.3018

a0 must be arbitrary but we have figured out that a1 is equal to zero, we are going to take this long expression and we are going to plug-in r= -1 into this expression .3037

We are going to try to get a recurrence relation to figure out higher coefficients in terms of lower coefficients.3051

Let me go ahead and plug-in r= -1 here, I get -1+3 xn - 1 + 2x n -1, now r= -1 is n + -1 -1, that is n -2, all of that times (an) + 2 a7 - 2 would have to be zero.3057

We are going to use this to get a recurrence relation that means you solve for the higher coefficient in terms of the lower ones.3090

I'm just going to do some algebra here -1+3n -3 + 2 x n^2 - 3n +2, Let me go ahead and keep solving that 3n -4+2n^2 -6n + 4.3098

My 4 is cancel that first 4 came from the -1 and the -3 by the way, 4 is cancel and so I get 2 1^2 -6n +3n, 2n^2 - 3n.3121

This was all multiplied by (an) and I'm going to move the 2an -2 over to the other side, so I will get negative 2(an) -2.3135

I'm trying to solve for (an) in terms of (an) -2 so my (an) is -2(an) -2/2n^2 - 3n so what that does is, it gives me a recurrence relation for the higher coefficients in terms of the lower coefficients.3149

Let me remind you that that is valid for all n bigger than or equal to, where I got that 2 from was right here, it is where the series started so that is where the recurrence relation starts.3171

This is really nice because now I have, I know that a0 is arbitrary I know that a1 is equal to zero and I have a way to find higher coefficients in terms of lower coefficients.3186

On the next slide, I'm going to go ahead and use this recurrence relation to figure out some coefficients of the series and write my generic series.3198

Let me recap briefly what we did on this slide, so from example 3 ,we already figured out the indicial equation and we figured out the roots r=1/2 and r= -1.3206

What were doing in this example is we are plugging in (an)r = -1, we plug it into this first term here, that is what we are doing right here.3222

We end up with is just an equation that degenerates into 0=0 which tells me nothing about a0.3236

a0 is arbitrary, when we plug r= -1 into this second term here it simplifies down to negative -a1=0 which tells me a1=0, so I figure out as much as I can about a0 and a1.3244

When we plugged r= -1 into this enormous term here, we plug-in r= -1 and it simplifies all the way down and then we try to solve for higher coefficients in terms of lower coefficients.3264

That sort of universal and series solutions, solve for higher coefficients in terms of lower coefficients so what we get is that (an) is -2(an) -2/2n^2-3.3280

We know that started in n=2 because of that being where the series started above, we are going to take these pieces of information, the arbitrary a0, the information on a1 and the information on (an).3294

And we are going to use those on the next slide to actually generate our series solution, let us go ahead and see how that works out.3309

Here is the recurrence relation (an)=-2(an) -2 times we had 2n^2 - 3n on the previous slide but here I just factored it into n x 2n - 3.3317

Let us go ahead and try plugging in different values of n and seeing where we get with that we know that, by the way let me remind you a0 was arbitrary, we figure that out on the previous slide.3331

a1 was equal to 0, we figured that out too , let us see what we can figure out if we plug-in n=2 here, we will get a2 is -2a0/ 2× 2n - 3 when n=2 would be 1.3349

If we plug in an n=3, we will get a3 = you know what it is going to be something in terms of a1, but I know that a=0.3371

It is just going to come down to be zero and in fact that kind of looking forward I can see these things are going to go by 2's, it is going to be odd terms, there is going to be even terms.3384

Because a1 was equal to zero, all the odd terms are going to be zero, I'm not even going to worry about those anymore, they are all going to be zero .3393

I have to keep going with the even ones so let us see what happens with n=4 plug in n=4 to the recurrence relation, I will get a4 is equal to -2a2/4, 2 x 4 - 3 is 8 - 3 is 5.3403

If I combine that up that was in terms of a2, but I have a2 in terms of a0 up here so if I plug in what I know about a0, about a2 there, I see a -2 x -2, so that is 2x^2 x a0/1 x 2 x 4 x 5.3426

I'm going to keep the even numbers together, I'm going to keep 2 and 4 together and 1 and 5 together.3446

Next term will be n=5, I already figured out that all the odd ones are zero so I can worry about that, we go ahead and plug in a=n=6 so I get my a6 is -2 a4/6 x 2 x 6 - 3 that is 2 x 6 is 12-3 is 9, 6×9.3454

I figured out a4 in terms of a0 so I see I have its this one is negative now, -2^3 x a0/2 x4 x 6 x1 x 5 x9.3483

I'm starting see a pattern developing here my a8 will be 2^4 x a0/ 2 x 4 x 6 x 8, 1 x 5 x 9 times next one they are going up by four, that would be 13.3501

I can simplify these bit, If go back to n=4 or even n=2, I can write that as -a0, just -a0/.3522

If I look at these in a4, I got 2^2 in the denominator and 2 x 4, if I cancel a little bit I got a0/ 1×2×1×5 and then with n=6 MIA 6.3535

If I cancel 2^3 in the numerator with 2×4×6 in the denominator, I got - a0/1 × 2 × 3 × 1 × 5 × 9 and finally my a8 2^4/2 x 4 x 6 x 8, that would give me 1 x 2 x 3 x 4 in the denominator.3555

That is just a0/4 factorial x 1 × 5 × 9 × 13, it is starting to look a little bit better and now let me assemble these pieces into a series.3579

Let me remind you what our original guess was, our original guess was y=x^r x the sum from n=0 to infinity of (an) x^n of course we combined x^r and got x^n + r.3595

I'm going to leave it separated for now, so my y=x^r and if I expand out that series it is a0 + a1 + a1(x) + a2(x^2) + a3(x^3) + a4(x^4) and so on.3614

In this case I have x^-1 now it is a0 so I do not know what a0 is, it is arbitrary such as write that is a0, a1 is zero's I can drop that term out and my a2 is -a0 - a0x^2.3641

My a3 is 0 because that is an odd one but my a4 is plus a0/1×2×1×5 all right that is 2 factorial x 1 x5 and that is all multiplied by x^4.3666

My a6 is right here, -a0/ 3 factorial x 1×5×9 x x^6 and my a8 is a0/ 4 factorial x 1× 5 × 9 × 13 x^8 and it keeps going like that.3683

I'm going to try to write a pattern for that it would not be the prettiest pattern in the world but I'm going to factor out an a0.3714

a0 x x^-1 and now I'm going to write a series and I see that in all of these I got an n factorial in the denominator is actually using a new version of (n).3724

I see I got an x^2n in the numerator because those are going up by even powers and then in the denominator, I see I have 1×5×9 and this seems to go up too.3741

Let us see, when (n) was 2, I went up to 5, when n was 3, I went up to 9.3757

They are going up by 4, that means it is a 4n, 4n when n = 2, that would be 2^8 so to get to 5, I had to subtract 3 here so (n) factorial times 1×5×9 up to 4n -3.3764

I can start that at n=0 to infinity if you write out this series you will see different terms the series, you will see that exactly the series above so my one solution here the a0 is just an arbitrary constant.3784

I'm going to leave that off and I'm going to combine this X minus one in with the x^2n so I'm going to get the sum from n=0 to infinity of x^2n -1/ n factorial times 1×5×9 up to 4n -3.3804

And that finally is my one series solution to the differential equation.3830

Let me remind you here, this was the series solution that came out of one of the roots of the indicial equation.3838

The indicial equation had 2 roots we had r=1/2 and r= -1, this is the series solution that came out of using r= -1, let me put a little reminder here.3846

Note that r=1/2 we could essentially do all the same work for r=1/2 and it would lead us down to a different series solution.3870

We are really only half finished this problem, we have to go through with r=1/2 and work through all this similar kind of work here and at the end we will get another solution with r=1/2.3902

That would be our second series solution to the differential equation, let me recap what we did here, this recurrence relation is what we figured out on the previous slide.3916

We went to a lot of work to figure out this recurrence relation, when we figured out that it is valid for (an) greater than or equal to 2, that is when we start plug-in values at n =2, n=3, n=4.3928

n=2 gave us a2 in terms of a0, n=3 gave us a3 in terms of a1 but we already figured out that a1 was 0 and so that means a3 is 0.3941

We see that it got to keep going, n=5 will give us a5 in terms of a3 which would still be 0.3956

All the odd ones are going to be zero, all the odd terms dropped out , but the inner ones keep going so n=4 gave us a4 back in terms of a0, n=6 gave us a6 back in terms of a0 and so on3962

We get a8 back in terms of a0 and the numbers we get in the denominator simplify a little bit down to this factorial pattern in the denominator.3978

I see that I missed one term when I was writing things out, I missed these negatives so I should have included that so let me go ahead and right that in now.3988

I will write it in green just so you recognize it there should be a -1^n here to keep track of the fact that these terms are alternating.4000

Let me add that in -1^n to keep track of the fact that these terms are alternating, just to keep going here we went back to our original expression.4009

This was our original expression our guess way back at the beginning of the problem x^r times our old series so I wrote in x^r here and I expanded out our old series a0 + a1x + a2(x^2) and so on.4021

I filled in what I knew about each one of these coefficients, I knew all the odd ones are zero, all the odd ones dropped out right away.4040

I filled in my coefficients a2 in terms of a0, a4 in terms of a0, a6 in terms of a0, a8 terms of a0.4048

I tried to find a nice closed form that describes those coefficients so I saw okay the powers on x are going up by 2, that is x^2n.4060

In the bottom I have factorial's and this 1, 5, 9 pattern which is a little ugly but I see it goes up by 4.4068

1, 5, 9, 13 it is going up by 4 that is y know it is 4n something and then I just kind of check a couple values of an to make sure that it was 4n -3.4075

For example when n was 2, 4n -3 is 5, when n is 3, 4n - 3 is 9 and so on.4087

n=4, 4n-3 is 13, I got that -3 in the answer there and then I dropped off the arbitrary constant just because I'm just looking for one solution and I combined this x^-1 in here.4095

I got x^2n -1 and finally I got this rather complicated series solution that was all applying to the root of the indicial equation r= -1.4113

There would be another whole series solution corresponding to r=1/2 which we have not worked out, that would be another long exercise to workout and get that second solution4127

That is the end of that problem.4139

In example 5 we have to find and solve the indicial equation for another differential equation and see what the roots are and determine which roots would lead to a valid solution4141

Let us go ahead and see how that one works out, again we start with the same exact guess as before that is y=the sum from n =0, (an) x^n since we are using series solutions around regular singular points.4153

We are going to jack that up to x^n + r and let us look at the different terms that were going to have to study here.4171

We are going to have a 3y here so I will go ahead and figure out that 3y is just the sum from an equal zero of 3an(x^n) + r.4181

And then we are going to have an x x y. so x x y is the sum from n=0 that just bumps it up by power of x, so (an) x^n + r +1 and now let us figure out Y′.4200

Y′ is the sum from n=0 of n + r x x x^n + r -1 and were going to have to deal with 3x Y′ and x^2 Y′.4222

Let me go ahead and work those out 3x Y′ is the sum actually that is going to be negative so I will go ahead and include that negative.4244

-3x Y′ would be, I will put the negative inside the series n=0 to infinity of -3n + r, a(sub)n, x would bump up the power of x^n + r.4254

Also, I have to deal with a -x^2 Y′ so -x^2 Y′ would be the sum from n=0 of -n + r(an)4281

The x^2 is going to bump it up from x^n + r -1, 2x^n + r +1, that is our -x^2 Y′ and finally we got a Y″ so let me calculate that, Y″ is the sum from n=0 of n + r.4298

We got a second derivative here, so I will ago n + r -1(an) x x^n + r - 2 but we are going to have to deal with x^2 Y″, I will go ahead and multiply that by x^2.4320

x^2 Y″ is the sum from n=0 of n + r x -1(an) and now the power gets bumped up to x^n + r, it was n + r - 2 before we multiplied by x^2 so it is x^n + r.4339

If you look at all these different series that we have to combine, remember there is a two-step procedure to combining series.4363

The first step is to match exponents, we want all those series in terms of x^n + r and the second step will be to match indices starting indices.4370

We will only get that after we match the exponents, we want the starting index everywhere to be the same.4393

Let us see what that has to be after we match the exponents, when I look at all these different series, I got x^n + r here that looks pretty good.4409

My xy has x^n + r + 1 and I would like to lower that down to be x^n + r, the way do that is you raise the starting index and correspondingly lower the n's in the formula.4418

I will go -1 on the n's in the formula +1 on the starting index and I will get the sum from n=1, I got to lower the n's in the formula, (an) -1 x^n + r.4437

Let us look at my 3x Y′ looks pretty good I got x^n there, that is looking good but my x^2 Y′ has an x^n + r + I got a fix that one, fix it the same way.4459

I want to lower that n by 1 which means I have to raise that n by 1, let me go ahead and do that.4476

I have the sum of n=1, I got to lower all of these n's, so -n - n + r -1, an-1 and I can lower the exponent x^n + r.4487

Finally, when I look at Y″ my x^2 Y″ that got an x^n + r, that is pretty good.4506

I got this series, I got this series, got this series, this series, and this series, all over now x^n + r, I have matched my exponents on all of them.4514

The next step is to match my starting indices which means I had to look at each one of these, I got n=0 here, n=1, n=0, n=1 and n=0.4528

I'm going to match all of the highest one which is n=1, that means for all the n=0 ones, I'm going to pull out one term.4542

I'm going to pull out the n=0 term, let us go ahead and start here, the n=0 term here is 3a0, x^r and that was my n=0.4554

Now I have the sum from n=1 of 3(an)x^n + r, this one is already okay because it already started at n=1.4574

This one starts at n=0 so I'm going to pull out the n=0 term which is -3 n=0, 3r(a0) x^r and then I can go ahead and write the rest of the series -3n + r (an)x^n + r.4585

I can start that at n =1 because can I pull out the n=0 term, this one is already ok because it starts at n=1 this one is not because it starts at n=0.4610

I'm going to pull out the n=0 term, pull out the n=0 term gives me just r x n=0, r -1 times (a0)x^r and then I will have the sum from n=1 of the rest of it which I'm not can bother to write because I'm running out of room there.4623

I have to combine all of these series so let me just indicate which of the series I have to combine, I have to combine that and that, and that ,and that, and this.4657

Let us see how that plays out I'm going to write all the extra stuff on the outside first, so I noticed that all these extra terms there is one there is one and there is one.4684

All of those have an a0 x x^r, now let us see what we have multiplied by that, that three is coming from here this gives us a -3r and then this gives us a plus r x r -1.4696

The other terms will all be the sum from and equals one and they all have an x^n + r, that will be an xp^n + r on all those.4725

It has not even been a bother to write down what the coefficient is because it be collecting a lot of messy terms is a good chance I will be make a mistake doing that.4735

I'm just going to leave that open right, the reason I'm not going to bother is because what the problem is asking me, the problem is just asking me to find and solve the indicial equation.4745

The indicial equation comes from this opening term right here, this gives me the indicial equation.4756

We are going to find that, we are going to solve it, and we are going to see which values of r would lead to a valid solution.4770

Let us keep going with that on the next page, the equation that we had right at the end of the previous page was 3 - 3r + r x r -1 x a0 x that is the coefficient of x^r + some stuff starting at n=1 x x^n + r + 1=0.4776

The x^r term tells us that it is coefficient must be zero,it tells us that 3 - 3r + r x-1 all times a0=0.4814

Remember an assumption we made for all of these regular singular points was that a0 is not equal to zero, we all assumed that a0 is not equal to zero that means that the other part here must be 0, we get 3-3r + r x r - 1=0.4832

That is the initial equation right there, but we are also asked to solve that and will go ahead and solve that, it should be a pretty easy quadratic polynomial, but if we multiply that out, we got 3-3r + r^2 - r =0.4855

r^2 -4r + 3=0 and if we factor that, that is r -1 x 3 -3 =0 ,so r= 1 and 3 we get two roots of the indicial equation and the problem asks us to determine which roots would lead to a valid solution.4882

The important thing here if you remember back to the very beginning, the lesson overview there is an important distinction when the two roots differ by an integer.4907

Here 3-1 is 2 and 2 is an integer so the difference between the two roots is an integer so since the roots differ by an integer.4920

What I said at the beginning of the lesson and the lesson overview if they do differ by an integer you can only make a solution out of the larger root, only the larger root r=3 would lead to a valid solution.4950

That is the end of what this example is asking us to do but if you wanted to keep going and finish solving this differential equation what you would do is you have to take r=3 and plug that back into this differential equation.4986

Get a recurrence relation and then build up your series solution one coefficient at a time it is rather messy and I'm not going to do it but that is how you would keep receiving here is with r=3.5000

Let me remind you exactly how we got to this stage, we started out with our generic guess y=the sum from n=0 to infinity of (an)x^n + r, we figured out Y′ and Y″.5013

Then we multiply Y′, y and Y″ by the various coefficients here x^2, 3x, x + 3, and so on.5036

We plugged those terms into what we got these big series and then we got all these different series the first thing we did was we try to match exponents,x^n + r and then we tried to match starting indices.5048

Now with the series, some of them started at n=0 and some of them started at n=1 so what we did was we took the n=0 series and we pulled out the n=0 terms and we started the series at n=1.5078

They all match at n =1, but then this n=0 terms those with the terms that gave us this extra equation out here, that came from the n=0 terms and then with those terms we found the indicial equation.5095

That turned out to be a quadratic polynomial that was using the fact that a0 is nonzero so that turned out to be a quadratic polynomial which is pretty easy to simplify, to factor and then solve in r = 1 and 3.5119

We get these two roots and we check their difference meaning we subtract them, we notice that their difference is not integer that the difference between the two is a whole number.5133

We had a rule from the beginning of the lesson that the roots differ by an integer, you are only allowed to use the larger one to get a solution .5143

Let us go ahead and finish this differential equation is you take that larger root r=3, plug it back into the recurrence relation and start solving for your coefficients one at a time.5152

That is the end of this lecture on regular singular points as part of the series solutions of differential equations. My name is Will Murray and you are watching www.educator.com, thanks for joining us.5166

Hi, welcome back to www.educator.com, my name is Will Murray and this is the differential equations lectures.0000

Today we are going to learn about the Laplace transforms, let us start with the definition, the Laplace transform of a function, so will write the function in terms of t.0006

The Laplace transform by definition that is this calc and equal sign means, its definition is the integral from zero to infinity of each of the negative st x f(t)dt.0017

Let me emphasize right away here the variable of integration here is t, I will be integrating in terms of t and then we are plugging in t=0 and we will take the limit as t goes to infinity.0032

That s never gets anything plugged in to it, so your answer will be a function of s, all the Laplace transforms that we do today will start out with a function of t, we will run the Laplace transform and we will end up with a function of s.0047

One really nice property about the Laplace transform is that it is linear which means that if we have a function f(t) and g(t) and constants a and b ,then the Laplace transform of a(f) + b(g).0064

You can just do the Laplace transform separately, you can do L(f) and you can do L(g) and then you can just carry along the constants on the outside.0078

That is very convenient, that means you can break up a function into its little pieces to the Laplace transform on each one.0087

Just put them back together it is very safe to do that, let us try taking some Laplace transforms and see what kinds of things we end up with.0094

The first example here is to find the Laplace transform of t^n where n is greater than or equal to zero, this is actually several problems in one because we want to find the Laplace transform for each different power of t.0104

After we do, we will know how to find the Laplace transform of any polynomial, so let us start by picking low values of n's, we will start out with n=0 here and that just means that the f(t) is equal to t^0 that is just 1.0119

We are going to take the Laplace transform of one, let me just remind you about the formula for the Laplace transform L(f), remember by definition is equal to the integral from zero to infinity of e^-st x f(t) dt.0137

When f(t)=1, L(1) is by definition is the integral from zero to infinity of e^-st and f(t) is just 1, so it is just dt.0160

We have to integrate that remember we are integrating with respect to t the integral of e^-st is just e^-st and then we divide by the coefficient of t so that is -1/s and then we integrate that from t=0 to what we take the limit as t goes to infinity.0177

Now when t goes to infinity remember this is a negative power of e, this is like saying 1/e to infinity that would just be zero, that is the the term when t goes to infinity, we would just give a 0 because we will have an e in the denominator.0206

When t=0 this is minus -1/s +1/s e^0 which of course is just one and so what we get for Laplace transform is 1/s.0224

That is our Laplace transform of 1and now let us do our next power up is t, L(t) again by definition is the integral from zero to infinity of F(t) x e^=st, that is t x e^-s(t) d(t).0239

In order to do that we have to integrate by parts and if you do not remember how to do integration by parts we have another whole series of lectures from the calculus 2 series here on www.educator.com.0267

You might want go check out the calculus 2 lectures here on www.educator.com and there is a whole lecture on integration by parts.0280

Today I will just use the quick shorthand version of integration by parts where you make a little chart here t and e^-st.0288

The quick shorthand version of integration by parts, write down derivatives on the left, t goes down 1, 0 t goes to derivative of 1 is 0 integrals of e^-st, that is -1/s e^-st and +1/s^2 e^-st.0300

That is the integral of the line above and then we write these diagonal lines and little signs plus and minus on the diagonal lines, this is the shorthand version of integration by parts.0322

We multiply along the diagonal lines and we get minus t/s e^-st -1/s^2, e^-st and then we have to evaluate this from t=0 to the limit as t goes to infinity.0336

We are going to plug in t going to infinity into each of these terms and what we see was, we will have an infinite term in the denominator there but then we have an infinite term in the numerator with that t.0365

And an infinite term in the denominator with e^-st but a negative exponent on the e beats any polynomial, so this term going to zero beats this term going to infinity.0380

If you want you can check this using L'Hopitals rule, that is how you could confirm this but since that is a calculus 1 topic, we are just going to assume it for here.0396

The infinity terms give you 0 - 0 and then zero terms give us while because when t=0 so this was t going to infinity and now we will do t as 0, this has t=0, there is a zero there.0407

And then +1/s^2 e^0 and that all simplifies down to just 1/s^2 because e^0 is 1, that means that the Laplace transform of t is 1/s^2.0428

Let us do one more here, this is when n=1, we are figuring out the Laplace transform of t^1, when n=2 we want to find the Laplace transform of t^2, again by definition that is the integral from zero to infinity.0448

t^2(^-st)dt and again we can use integration by parts on that, and integration by parts I'm going to use the shorthand version.0471

You can go back and check the calculus 2 lectures are on www.educator.com if you are a little rusty on your integration by parts.0484

I'm going to write derivatives first derivative is 2t, derivative of 2t is 0, derivative of 2 is 0, integral of e^-st is -1/s, e^-st the integral of that is positive 1/s^2, e^-st.0492

The integral of that is -1/s^2 e^-st and I will going to write my diagonal lines again and put my alternating signs plus minus plus and I'm going to multiply along the diagonal lines.0512

I get -t^2/s e^-st - 2/s^2 e^-st - 2s^3 e ^-st and again I have to evaluate that from t=0 the limit as t goes to infinity.0530

I have e^-st on every term so even though that is getting multiplied by some positive powers of t having an e^-st in the denominator will drag all of these terms down to zero when t goes to infinity.0561

This is 0-0-0 when t goes to infinity, I'm kind of slurring over what is really L'Hopitals rule so if you are not so sure why these all go to zero.0577

Just check them out using L'Hopitals rule which is something you can learn on www.educator.com from the calculus 1 lectures and now if we plug in t=0, first term here has a t^2 in the denominator so that is 0.0590

Second term has a 2t in the denominator so that is zero, third term is 2/s^3 x e^0 by way, I have been putting plus on each of these because t=0 is the lower limits.0605

We are actually subtracting these off but then we had these negative signs so we are subtracting a negative which makes it positive and this gives us e^0 is 1, 2/s^3 is our Laplace transform for t^2.0620

We are starting to notice a pattern here, we are going to take one more power of t and then we are going to figure out this pattern for sure.0640

Let us try n=3 so the Laplace transform of t^3 is the integral from zero to infinity of t^3 e^-st(dt) and we are going to use parts again.0647

t^3 e^-st and again I'm going to use my shorthand version of integration by parts so derivative of t^3 is 3t^2, derivative of that is 6t, derivative of that is 6 and derivative of that is 0.0671

Integral of e^-st is -1/s e^-st, integral of that is 1/s^2 e^-st, integral of that is -1s^3 e^-st,the integral of that is 1/s^4 e^-st.0694

Now were going to write diagonal lines on each term here and there is a plus minus plus minus and so we can write down what our answer is.0724

This is the tabular integration method of integration by parts so we get minus t^3/s e^-st -3t^2/s^2 e^-st - 6t/s^3 e^-st and -6/t^4 e^-st and all of this has to be evaluated from t=0 2t going towards infinity.0739

Once again, if we take the limit as t goes to infinity of each of these terms , we will have some infinities in the numerator but nothing strong enough to be the e^-st which gives us infinity in the denominator.0783

All the infinity terms will give us zero and the t=0 terms, most of them will be 0 too, because we got t^3 and then t^2 and then a t, but then we have 6/ - that should have been s^4 not a t^4 up there.0798

We have plus because we are subtracting a negative, 6^4 e^-st, let me write that a little more clearly, 6/s^4 e^-st and so those are all the terms we get by plugging in t=0.0828

Actually when t=0 e^-st just gives us e^0, e^0 is 1 so it just simplifies down to 6/s^4 and that is the Laplace transform of t^3, that is what we started out with.0855

I think it is time to start noticing a pattern here what you can notice here is that 6 came from this 6 over here originally and this 6 came from 3× 2.0872

What is going to happen when you take higher powers is are going to be multiplying on bigger and bigger numbers and that 6 really comes from 3 factorial/s^4.0885

What we notice here is that the Laplace transform of one was 1/s, that is what we figured out on the previous page.0899

Laplace transform of t was 1/s^2 Laplace transform of t^2 is 2/s^3 we are starting to build up this factorial pattern in the numerator.0909

Laplace transform of t^3 is 6, 2×3/s^4 and that is because we are building up that factorial pattern in the numerator so in general our Laplace transform of t^n is n factorial.0931

We are getting an n factorial and in the denominator, when we had t^3, we had s^4, when we had t^2, we had s^3, each time we are getting s^ n +1 so that is our general formula for the Laplace transform of t^n.0961

Let me recap what we did there, basically what we did here was we had several different problems because we are trying to figure out the the Laplace transform of t^n for all values of n bigger than zero.0983

What we did was we took the n's one at a time, each time we took a value of n plugged in t^n here and that meant we took the integral of t^n x e^-st.0998

That is coming directly from our original definition of Laplace transform which was the integral of e^-st x f(t)dt, evaluated from zero to infinity.1009

We worked out that integral but in all of these we had to do integration by parts and I did not really show you the details of the integration by parts.1023

I use this short hand tabular integration method that we learned back in calculus 2 to do the integration by parts.1031

We worked out the integration by parts and then we try to plug in t going to infinity and we figured out that since we had a negative power of e on each of these terms, that gives us an infinity in the denominator.1039

That is stronger than any of the infinities in the numerator, that is kind of short hand way of getting around L'Hopitals rule but to check it formally and confirm it you would run L'Hopital's rule that shows you the all the infinity terms go to zero.1054

We plug in t=0, most of these terms still go to zero because when t=0 there were 0 in the numerator of most of these terms but this last term did not have a t in the numerator.1068

We got 6/s^4 x e^0 which is where we got our Laplace transform for t^3.1082

We did this for each power of t until we noticed a pattern, and the pattern is that we had this factorial these factorial is building up in the numerator, we always had s +1 in the denominator .1090

That is our generic formula for the Laplace transform of t^n is n factorial/s^n +1, in our next example we have to find the Laplace transform of f(t) equal e^(at) assuming that s is bigger than a.1101

Let us work that out, again from the definition of Laplace transform, let me mention first that s bigger than a, that tells us that s - a is going to be a positive number and that is going to be useful as we do our integration.1120

Let us remember the definition of Laplace transform L(e^at) by definition of Laplace transforms, this is using our original definition, is the integral from zero to infinity of e^-st times whatever your f(t) is.1139

In this case, it is e^(at)dt and what I'm going to do is I like to combine these exponents the (st) and the (at) and I'm going to factor out the negative sign.1157

We have the integral from zero to infinity of e^negative, now this is s - a, because I factored out a negative sign times (t)dt, the key point there is that s - a is positive.1171

We figure that out at the beginning using our assumption, if we integrate that, that is e^ negative - (at)/(-s - a) and we are going to evaluate that from t=0 to the limit as t goes to infinity.1188

That means were going be plugging in t going to infinity in here, but remember s minus a is positive, that is what we set up here s minus a is positive.1213

e^s - a is a negative power of e, when t goes to infinity we got a negative power of e, that means it is going to 0 and when t equal zero, were subtracting the t=0 term and that is negative so it turns into a positive.1222

We get e^0 because we are plugging in t=0/s -a r and it simplifies down to 1/s - a, remember e^0 is 1, we are done with that one.1243

Let me recap what we did, we started out the original definition of Laplace transform, that is the integral from zero to infinity of e^-st times whatever the f(t) is.1262

We took that f(t) and drop that into that integral formula, then I combined the exponents into s - a while negative s - a.1273

What I'm really doing is changing that positive a there into a (-a), it turns out to be convenient when we do the integration.1286

The integral of e^-s -at is just the same thing divided by negative s - a and then we plug in t goes to infinity so that is a negative power of e.1296

That is why we got zero here, it is e^negative infinity which is the same, remember 1/e to the infinity and then t=0 because we had t here that gives us e^0, that is just 1/s - a.1308

On our next example, we are going to final the Laplace transform of f(t)= cos(at), we are going right from the definition of Laplace transforms.1327

The Laplace transform of cos(at) just by definition is the integral from zero to infinity of e^-st times whatever your function is.1338

In this case, its cos(at) dt and this integral would be kind of ugly if you encounter this in calculus two class, you would probably use integration by parts twice.1354

Integrate by parts twice, alternately you could use a computer algebra system or an online integration tool, you can use that to solve this integral.1373

Or you could use a chart of integrals which are probably find inside the back cover of your calculus book there is lots of charts of integrals there and this will definitely be one of them.1393

Because it is a little cumbersome I do not want to go through the details here, I'm just going to read the answer off, I got this from a chart of integrals but you could also find it by any of these other methods.1408

It is negative s/a^2 + s^2, I'm reading the integral of e^-st x cos(at) tuns out to be -s/s^2 + a^2 x e^-st cos(at) + a/a^2 + a^2, e^-st sin(at).1420

What we are supposed to do with this is evaluate it from t=0 to the limit as t goes to infinity.1455

We got to plug our limits t going to infinity and t=0 in here, what we notice is that we have an e^-st, remember that is going to have an e in the denominator, whenever we plug in t=infinity or t approaching infinity this will be 1/e^infinity.1478

Will this be one to the infinity and so both of those terms go away to zero and when we plug in t=0 the sin term is going to be zero, let us remember that sin(0) is equal to zero and cos(0) is equal to 1.1502

The sin term goes to zero the cos term goes to 1, so minus s/a^2 + s^2 x e^0, that is when we plug in t=0 and it is minus all of these because it is the lower limit here.1534

All these terms dropout except for this term that e^0 is 1 and the two negatives cancel so we get s/a^2 + s^2 and it is positive because the two negatives cancel each other, that is our Laplace transform of cos(at).1556

We get a function of s there should not be any t's left over in these when you are done with taking the Laplace transform so let me recap how we got that.1577

We use the straight definition of Laplace transform which says you do the integral of e^-st times whatever your function is, we dropped in cos(at) now that is the integral that would have been a bit of a headache in calculus 2.1588

One way to do it is to do integration by parts twice, that is if you are going to do it by hand.1601

Another way do it is just drop it into a computer algebra system that will work out for you or an online integration system which there are several these days.1608

You can also use an integral chart if you still got your calculus book just check in the back cover you will see all kinds of charts of integrals and this is will be one of you them.1616

e^-st x cos(at) you might have slightly different variables, it might have a and b, it might have x instead of t but basically you will see that integral.1624

And you will see it expanded out into -s/a^2 + s^2 e^-st cos(at) and then a/a^2 + s^2 e^sc sin(at).1635

We try to plug in the limit as t goes to infinity everywhere, but when we do that, these e terms we had negative exponents on the e.1650

We get each of the infinity in the denominators all the all those terms turn into zero and dropout.1660

It is just a matter plugging in t=0 when that goes into sin we just get zero here so that term drops out.1667

We plugged in the cos(0) is 1 and we have to keep that term, we get an e^0, that gives us 1 and finally we just get s/a^2 + s^2, of course it is negative from this negative right here, that came out of doing the integral.1675

Since it is the lower limits, we have another negative sign here and so those two negatives gives us a positive.1694

We are finally just left with s/a^2 + s^2, so example 4 is quite similar to example 3 will do it in kind of a similar fashion, we are going to find the Laplace transform of f(t) is sin(at)1702

We are going to use just the straight definition of the Laplace transform L of sin(at) by definition were using the definition of Laplace transform.1717

Integral from zero infinity of e^-st times whatever function you are talking about, that is the sin(at) in this case dt.1729

This is kind of a nasty integral, just like in example 3, we would use parts integration by parts twice or we would use a computer algebra system or an online integration system or an integral chart.1743

Any of these other methods should give you the answer to this integral, what it should tell you is -a/a^2 + s^2, e^-stx cos(at) + s/a^2 + a^2 x e^-st x sin(at).1775

We have to evaluate that from t=0 to the limit as t goes to infinity, let us look at what happens when we plug in t=infinity.1808

Here we got a negative exponent to the e, I wil have e^infinity, which is the same as 1/e^infinity and we have the same thing over here.1827

That means when t goes to infinity, we will have these two terms that both go to zero.1837

When we plug in t=0, the first term is cos(at), that is cos(0), we will have -a/a^2 + s^2, e^0 x cos(0) + s/a^2 + s^2 e^0 x sin(0).1848

That is what we get plugging in t=0, the sin(0) goes to 0, that means this entire term drops out,cos(0) gives you 1, e^0 gives you 1, and we have -a/s^2 + s^2.1878

Those two negatives cancel and gives us a positive a/ a^2 + s^2.1896

We are done with that Laplace transform, let us recap and see how that worked out.1910

We started out with the definition of Laplace transform, that means e^-st times whatever function you have x f(t), we dropped that and integrate that (dt).1914

The integration is messy because it got an e in the sin term, if you did that in a calculus 2 class, you would be doing integration by parts twice and then solving around for the original function.1926

You could also throw the whole thing into a computer algebra system software package or an online integration tool, there are several of those that you can use.1939

Or you can look at the integral chart which is probably located on the inside back cover or maybe front cover of calculus textbook, almost all of them have integral charts.1948

What you will see is a formula for e^-st sin(at) and when you integrate, turns into this complicated expression with e^-st(c0s) e^-st (sin).1958

We want to evaluate that from t=0 to t going to infinity,the top term going to infinity turns into 1/e^infinity for each of these terms, each of those is 1/e^infinity, which goes to 0.1972

All the infinity terms dropped out to 0,when you plug in t=0 we got cos(0) is 1, e^0 is 1, you will get -a/a^2 + s^2, plug in t=0 on the other side we get sin(0)=0, that whole term drops out.1989

We have minus, this is minus because we are subtracting the two limits, t goes to infinity - t goes to 0, that is coming from there.2006

This negative came form that negative right here, which just comes out of what the integral chart gives you as the answer to the integral.2018

These 2 negatives cancel each other out and we end up with a/a^2 + s^2.2027

On our final example, we have a very complicated function f(t) is 3(cos) 4t - 2(sin) 5(t) + e^2t + 3t^2 + 7t -2.2034

We want to find the Laplace transform for that, if we use the definition of the Laplace transform, by definition we have to do the integral form 0 to infinity of e^-st x f(t)dt.2047

Our f(t) would be this enormous expression, we have to plug all of that into f(t) and we just get this horrific integral and we really do not want to do that.2066

There is a much better way, we are going to use linearity and we are going to exploit the fact that we already know the Laplace transform of the various pieces of this function r.2078

This linearity of the Laplace transform is going to be extremely useful here.2093

We already know what the various pieces of the function do when you take their Laplace transform, L(1) we figured this out, this is back in example 1.2099

You might want to go back and check at example 1 if you do not remember this, it was 1/s, L(t) was 1/s^2.2109

L(t^2) was 2/s^3, I'm looking at the various pieces of this function here and I'm trying to remember what the Laplace transform of each one was.2128

These came from example 1, if it is been a while since you worked through example 1, maybe go back and take a peek at that and you will see where this come from.2144

L(e^at) is 1/s-a, I believe that was example 2, you might want to go back and check that out if that does not look familiar.2154

L(cos) of (at) worked that one out in example 3, and that was s/ a^2 + s^2 and finally L(sin)(at) was a/a^2 + s^2.2177

We worked that one out in example 4, we are not really going to do any new math in this example.2208

We are just going to exploit all the work we did in all the previous examples and that should be pretty quick.2218

I'm going to look at this f(t), I see 3(cos)4t, here is my Laplace transform for cos(at), the Laplace transform of f is from the 3(cos)4t, I'm going to get 3.2224

cos(4t) is going to give me s/16 + s^2, because the a is 4 there, -2 x sin(5t), sin gives me a/a^2 + s^2, a is 5 so 5/25 + s^2 + e^2t.2247

I can read that one right here, that is plus 1/s - 2 + 3t^2 + 3, t^2 gives me 2/s^3, 3 x 2/s^3 + 7t + 7 x 1/s^2 - 2 x 1/s.2276

The Laplace transform of 1 is 1/s, maybe I can clean that up a little bit, that is 3s/s^2 + 16 -10/ s^2 + 25 + 1/s - 2 + 6/s^3 + 7/s^2 - 2/s.2304

That was my Laplace transform of this large complicated function here, let me recap how we worked that out.2347

We did not want to go back to the definition of the Laplace transform, that would involve writing the integral of e^-st x f(t), where f(t) is this enormous function.2353

We have been to a horrible integral to work out from scratch, we are going to use linearity and we are just going to break this function up to its pieces.2364

Each of these pieces is something out I figured out, the Laplace transform earlier.2371

Example, the first time I look at this polynomial, we have the basic pieces are the 1 and t and t^2, and we figured out the Laplace transform for each one of those in example 1.2377

Laplace transform of one is 1/s, t gives us 1/s^2, t^2 give us 2/s^3, I got the Laplace transform of each of those.2391

e^at in example 2, I figured out its Laplace transform is 1/s-a, for the cos and sin, I figured out the Laplace transform of those in example 3 and 4.2403

Those were s/a^2 + s^2 and a/a^2 + s^2, what I did was I took each of those functions and just plug them back in here and then attach the right coefficient.2416

3/ transform of 3 + cos(4t), my a here was four, that is where I get that 16 + s^2, for sin my a was 5, that is where I got that 5, that 25.2428

Plugged in that -2 as a coefficient here, e^2t my a was 2 there, plug that in there as 2 and I get 1 - s/2 and then t^2 gave us Laplace transform is 2/s^3.2449

There it is there, Laplace transform of t and the Laplace transform of 1, I just cleaned everything up, combine the 3 s, 2 x 5 is 10, combine 3 x 2 x is 6.2466

Combine everything together and finally I got that Laplace transform of that big, horrible function without ever actually having to do any new integration.2479

I just relied on what I had worked out in the previous examples.2488

That is the end of our lecture on Laplace transform, in the next lecture we are going to learn about inverse Laplace transforms, learn how to go backwards from the answer of a Laplace transform back to the original function.2494

That is our next lecture in the differential equation series here on www.educator.com. My name is Will Murray, thanks for watching.2505

Hi and welcome back to the differential equations lectures here on www.educator.com, my name is will Murray and today we are going to talk about inverse Laplace transforms.0000

The idea of the inverse Laplace transform is that we are given the answer after somebody else is done the inverse Laplace transform and were going to try to reconstruct the original function f(t).0010

We begin in a function in terms of s and we have to figure out which function in terms of t would have given that by taking the Laplace transform.0023

Usually what happens is the function will be given will be a rational expression in terms of s, that would be a polynomial and s divided by another polynomial in s.0035

It will be a rational expression in terms of s and what we want to do is we want to run partial fractions on it in order to separate it out into simpler expressions in terms of s.0059

Once we get into simpler expressions in terms of s, we are going to use this table of common Laplace transforms, these were all of Laplace transforms that we worked out in the previous lecture.0075

If this do not look familiar what you might want to do is go back to the previous lecture here on www.educator.com and you will see a lecture on Laplace transforms.0087

You will see where each one of these terms comes from, we figured out that if you start with one then what you get when you take the Laplace transform is 1/s, t gives you 1/s^2, t^2 gives you 2/s^3.0096

In general, the Laplace transform of t^n is n factorial/s^n +1 and then we figured out the Laplace transform of e^at is 1/s - a, te^(at) is 1/s - a^2 and in general t^n e^(at) is n factorial/s - a^n + 1.0110

We figured out the Laplace transform of cos(bt) is s/s^2 + b^2 sin(bt) gives you b/s^2 + b^2 and then you can combine e^(at) cos(bt) and e^(at)sin(bt) to get s - a/s -a^2 + b^2 or just b/s - a^2 + b^2.0137

The idea is you really want to have this chart handy because we are going to run partial fractions on the expressions we are given and then we are going to compare our answers to this chart.0162

We get answers in terms of things like s - a and s^2 + b^2 we get answers that look like that, and then we will come back and look at this chart, and we will say that must have been the Laplace transform of e^(at).0176

That must have been the Laplace transform of cos(bt), that is how this works.0190

Let us try that out on some examples and you will see how it goes, the first example is to find the inverse Laplace transform of 7s +5/s^2 + s -2.0197

The first step here is to run partial fractions on the expression we are given, now partial fractions is something that you really learn about in calculus 2.0213

If you do not remember how to do partial fractions, what you really want to do is go back and look at the calculus 2 lectures, we do have some calculus 2 lectures here on www.educator.com.0229

You will see there is a whole lecture on partial fractions where you can practice this technique a little bit before you start using it for inverse Laplace transforms.0242

What I'm about to do with partial fractions here is mysterious to you or you are rusty and have not done it in a couple years, maybe it is worth going back and watching that other lecture on partial fractions before you come and work through this one.0251

Let us go ahead and now I'm going to assume that you remember partial fractions, we are going to write 7S +5, I'm going to factor that denominator there, that s^2 + S -2.0265

I can factor that into S +2 times S -1 and then remember the way we do partial fractions is we write that as a/S +2 + B over S -1 and then you try to figure out what the a and b should be.0278

The way we do that is you clear the denominator, I'm going to multiply both sides by my common denominator and here my common denominator is S + 2 x S -1.0300

I get a common denominator and I get 7S + 5 is equal to a x S -1+ b x s + 2 and now want to figure out what A and B are and there is kind of a clever trick which is to plug in some good values of s.0313

The clever way to do that is to plug in S=1 and that will make that term dropout so s=1 and if I do that on the left-hand side, I will get 7s +5 that is 12 is equal to a x 0+ b x 3 and that tells me right away the b=4.0335

If I go back and if I plug-in S=-2 then 7 x s will be -14 + 5 is -9 ,we get -9 is equal to a x s -1 will be -3a, a x -3 + b x 0.0362

That is why I picked -2 was to make that b term dropout, I get -3a = -9, a=3, and my partial fraction expansion of 7s + 5s^2 + s -2 is my a was 3, that is 3/s + 2 + my b is 4/s - 1.0389

That is my partial fraction expansion, I have to remember the relevant entries from a chart on inverse Laplace transforms what I remember is that the Laplace transform of e^at is equal to 1/s - a.0422

This is coming from the chart of Laplace transforms, that was not something I figured out on the spot, that was something according from the chart.0443

I'm going to use linearity here, I see 3 x s + 2, f(t) that means my a must be -2, that is 3 e^-2t + 4 my a must be 1 now, 4e^t.0461

That is a function where we took its Laplace transform, you would get this function of s that we started, let me recap how we did that.0487

First thing we did here was we expanded, we factor the denominator into S +2/S -1 and then expanded it out into partial fractions form.0497

If you are rusty on your partial fractions that was really a calculus 2 topic, in the beginning of calculus 2 we learned about integration techniques, check back at your calculus 2 notes.0507

Maybe watch the calculus 2 lecture on partial fractions here on www.educator.com and you will see how we did this, kind of the details of that but we clear denominators here.0516

I got 7s + 5= a x s - 1 + b x s +2 and we wanted to solve for a and b, in order to make certain terms dropout, we can choose good values of s and plug it.0527

Here we plug in S=1, which makes that term go to 0 and we quickly get 12 = 3b, b=4, plug in s= -2 which makes the b term go to 0.0541

We can solve for the a term is equal to three, that tells me what the partial fraction expansion of our original function is and then I go back and I look at my chart of Laplace transforms and I try to find something that matches these expressions here.0554

What I see is the Laplace transform of e^at is 1/s - a , that matches both of these expressions and the first one a= -2 and I got 3e^-2t, second one a=1, I just got 4e^t.0571

That is my inverse Laplace transform of the function that we started with, let us try another one here, we are going to find the inverse Laplace transform of -2s -3 / s^2 + 3s.0589

All hinges on partial fractions, you really want to be solid on your partial fractions before you mess around with inverse Laplace transform, we are going to go -2s -3 and I'm going to factor the denominator again.0604

s x s + 3 and then set up my partial fraction expansion which is a/s + b/s + 3 and I'm going to clear my denominators by multiplying both sides by s x s + 3.0621

On the left I get -2s -3 is equal to a x s + 3 + b x s and now I want to figure out what a and b are.0644

To do that, I'm going to choose good values of s that can make terms dropout selectively, I see that if I plug in s=0 that make the b term dropout and on the left I get -3, on the right I will get a x 3 + b x 0.0660

That tells me right away that a is -1, if I plug in s= -3, I'm going to make the a term dropout, I will get a x 0 + b x -3.0679

On the left if I plug in -3, I will get -2 x s, that is -2 x -3 that is 6 -3 is + 3 and that tells me that my b is also equal to -1 and now that I know what my partial fraction expansion is, I get -2s -3 /s^2 + 3 s= -1/s - 1/s +3. 0696

The chart of Laplace transforms is what I'm going to look at, that was back at the beginning of this lecture.0735

We had a chart of Laplace transforms if you want to have that chart somewhere near you whenever you are looking for inverse Laplace transform.0743

You do not want to figure it out from scratch, you want to look at the chart and see which kinds of functions correspond to which Laplace transforms.0752

In this case, I notice that L(1) is 1/s, that one matches that one, and L(e^at) is 1/s - a, that one matches that one and so my f(t) in order to produce this Laplace transform must have been -1 to get the -1/ S -1 over, not one over but e.0762

Since its s +3 that is like s -3, my a is -3, e^-3t, that is the function that we must have started with if we took the Laplace transform and got this function of s that we are given.0798

Let us see how that works out, again we went partial fractions which means factoring the denominator, setting up the partial fraction form, clearing the denominators by multiplying by the common denominator.0815

We get this expression in terms of s, we are trying to solve for a and b but we can do that by plugging in clever values of s, we plug in s=0, I get a nice, easy equation to solve for a.0829

Plug in s=-3 that makes the a term dropout, we get a nice easy equation to solve for b and we now have this partial fraction expansion of the original function.0841

We look at that and we then we look back at our chart of Laplace transforms, try find something that matches it and I can find something, I see L(1) is 1/s, L(e^at) is 1/s - a.0854

I plug goes back in and I see that in the first one I just got -1, in the second one my a must be -3, I got e^ -3 t, I know that the original function there, the inverse Laplace transform of what we started with is -1 -e^-3t.0868

In the next example, we got find the inverse Laplace transform of s + 4/s^2 + 4s + 5, now I like to run partial fractions on this one which would mean factoring the denominator.0890

The problem is that denominator does not factor, there is no way to break this up using partial fractions instead what I'm going to do is another old algebraic technique and I'm going to complete a square on the denominator.0905

It does not factor if it did factor, I would not be bothering to complete a square, s^2 + 4s + 5= I'm looking at that s^2 + 4, remember how complete a square, you take the coefficient of s and divided by 2 and square it.0923

4/2 is 2 and 2^2 is 4, I write s^2 + 4s + 4 and then I see that in order to make this equal, I had to add another one here.0945

The whole point of that was that gave me a perfect square (s + 2)^2 + 1 and my expression here is now s + 4/(s + 2)^2 + 1.0962

Instead of running partial fractions I completed the square in the denominator and now I'm going to look back at my Laplace transform chart and see if I see anything that matches that.0983

Laplace transform chart and what I notice there is the Laplace transform of e^at cos (bt), according to the chart is s - a/(s - a)^2 +b^2 and L(at) sin(bt) is b/(s - a)^2 + b^2.0996

I got a couple terms here that seem to match what I have for my function pretty well but I see that in order to use this cosine term, I got to have that s - a matching what is in the square there.1043

I do not have that right away here and I have s + 4 not s +2, I'm going to write this as s+ 2/s +2^2 +1 + now since I have s + 4 before, I have to add on another 2/s + 2^2 + 1.1061

I look at that and this first one is exactly tailor-made for cosine transformation with a=-2 and b=1 and this second one is ready for assigned transformation again a= -2 and b=1.1087

There is a slight incongruence there because here we have b and here we have 2, but I will just write that as 2 x 1, we will think of that 2 as being a coefficient, I will put hat on the outside.1109

Our f(t) from that first term we rig that up to be a cosine transformation, our a was -2 so that is e^-2t x cos(t), the cosine of (bt) is 1 and our sine transformation on the other one is e^-2t x sin(t).1125

We had a 2 in the numerator and we are just supposed to have a b there and our b was 1, we put a 2 out here.1152

Just think of that 2 as a coefficient and that is the f(t) whose Laplace transform would give us that function of s that we started with.1158

Let us recap how we figure that out there, first thing we try to do is we are hoping to do partial fractions on this function we started, we are hoping to factor that denominator but it does not factor.1175

Instead of factoring it, we are going to complete the square, here is what I'm doing is completing the square here, I take 4/2 and I square it and I get 4, that is where that 4 comes from.1189

Since I had a 4 on the inside here, I had to have a 1 here to balance out this 5, I write this is as s + 2^2 +1, that is the first thing I do there is create this denominator S +2^2 +1 then I glanced at my Laplace transform chart.1205

You really want to do this all in your head, you probably want to be looking at Laplace transform chart while you are solving these problems and I look for something similar and I see e^at cos(bt) and e^at sin(bt).1224

Give me expressions that are quite similar to what I have but the cosine one, the sin one is just a constant, the cosine one has s - a which matches the s - a in the denominator.1240

I did not have that right here, I split the s + 4 up in the s + 2 and 2, now that S + 2 matches the denominator and it is perfect to set up the cosine transformation where my a is -2, my b is 1.1255

That is where I got this e^-2t cos(t), now the sin transformation, I see b/s - a^2 + b^2, because that b1 has to be 1, which means the two does not quite match.1273

If I write that two was a coefficient that 2 will just come down on the outside and then it is okay to use b=1, I get e^-2t x sin(t).1289

What we found there was a function that if you took its Laplace transform, you would have gotten the original function that we started with, what we found was the inverse Laplace transform of that function of s that we started with.1300

The next example we got find the inverse Laplace transform of 4s^2 -11 s +11/s - 1^3, once again this is kind crying out to run partial fractions on it, let us go ahead and set up the partial fractions here.1323

The denominators are already factored, that is quite convenient, I'm going to set up my partial fractions format for one I have cubed in the denominator, 4s^2 - 11s + 11/s - 1^3 is equal to- remember you go a/s - 1.1346

You can not just go +b/s -1 + c/ s -1, again because that would just combine into a + b + c/s - 1, just a single s - 1, you will never get S -1^3.1372

Here is what you do in partial fractions, you go b/s - 1^2 and c/ S -1^3, again I'm relying on the fact that you remember your partial fractions from calculus 2.1389

If you are little rusty on that, we got a lecture on partial fractions in the calculus 2 series over here on www.educator.com.1402

In the mean time and let us keep moving forward, we are going to clear our denominators here, that means multiplying everything by S -1^3.1410

That means on the left we get 4s^2 -11s +11= a now we got an s -1 in the denominator and s -1^3 in the numerator.1422

Lined up are just a x s -1^2 + b x s - 1 because to of the powers cancel, just +c as a constant, that is very nice there, we want to solve for a, b and c.1435

This is going to be a little bit messy but if we do something clever, we can plug in s=1, if we plug in s - 1 right away, on the left hand side we will get 4 -11+ 11, that is 4 = a on the right hand side a x 0 + b x 0 + c.1451

That will tell us very quickly that c=4, that is nice because it will mean that later on we will not be solving for three equations and three unknowns, we just are a and b now, we know what the c is.1478

Let us keep going with expanding that right hand side, we get 4s^2 -11s + 11 is equal to a x s -1^2 is s^2 - 2s + 1 + b x s - 1.1492

Now +c, I already figured out that is 4, this is (as)^2 - 2(as) + a + (bs) - b + 4 and if I look at this and start preparing coefficients from the s^2 term of both sides, I see that I get 4 is equal to a.1510

That is very nice I know immediately that a=4, Let me go ahead and use the s term, the s term on the left hand side will give me -11, on the right hand side will give me, I have s terms here and here.1538

-2(as) + b, I already figured out that a was equal to 4, -11 is equal to -2 x 4 is 8 + b, if I move that a over to the other side, I get b is equal to -11 + a is -3.1556

I figured out my a, b, and c here, that means I'm done with my partial fractions expansion and let me just summarize what we figured out just from our partial fractions from our algebra here.1585

We got for 4s^2 -11s +11/ s -1^3 is equal to a/S -1, there is my a right there is a = 4, 4/s -1 + b/s -1^2, I see that b is -3, I change that plus to minus -3 /s -1^2 and +c /s -1^3, my c I figured out was 4.1601

4/S -1^3 and I'm going to use my Laplace transform chart, let me copy a couple of entries from the Laplace transform chart which I think will be useful here.1642

This chart I posted it in one of the slides at the very beginning of this lectures so you can rewind and check that out if you do not remember where this coming from but I'm looking for terms that look like these ones that I have in my expression for S here.1671

What I see is that L(e^at) is 1/s - a, that is certainly will be useful L(te^at) is 1/s - a^2 and in general L(t^n) x e^at is n factorial over s - a^ n +1.1687

I see down here that I'm going to have s - a^3, that means I'm going to be dealing with L(t^2/e^at), let me go ahead and plug in an n = 2 here.1734

2 factorial is just 2/s -a^3 because it is n + 1, I'm going to compare these terms with the Laplace transform that I have been given, 4/ s -1, I'm looking up here at 1/s - a, this is 4 looks like a=1 on all of these.1749

This is 4e^t and then -3 x te^t because I'm working up here at 1/s - a^2 and now for the 4 x s -1^3, I see that up here I have 2/s - a^3, I going to write this as 2 x 2/ s -1 cubed.1781

This is + 2 x t^2 e^t, my a is one all those and that is the function for which if I took the Laplace transform I will get this rational expression that we started with, I will give you a quick recap of how we did that.1810

We started out with this nasty rational expression and we ran partial fractions on it, the partial fractions expansion when you have S -1^3 in the denominator this is set as s - 1, S -1^2 and S -1^3 .1834

Then we are clearing the denominators by multiplying both sides by S -1^3 here and that gave us 4s^2 -11s +11 is equal to a x s -1^2, that is canceling a couple powers of s -1 to get b x s - 1.1851

And cancelling all the s -1 to get the c there and then I noticed that if I plug-in a good value of s=1, I can make the a term and the b term dropout quickly figure out that c=4.1873

It gets a little harder after that, I expanded out my right hand side S -1^2 is S^2 -2s + 1 and then I distributed the a through all the terms here and I sorted out into powers of s.1886

I sorted out the s^2 term and the in the S terms, the s^2 is just a 4 here and an a over here ,I get 4=a, the S terms I see -11 over on the left and a -2a and a b on the right.1901

That is why I get the -2 a and b, I already figured out that a was 4, I plug that in, get 8 there and then I find out that the b is -3, that means I figured out my a was here, my b was here and my c was here.1921

I plug each one of those back into this generic partial fractions formula, got the a, b, and c now, that is where I got the 4 -3 and 4 there and now I want to look back at my chart of Laplace transforms.1938

And see if I can find something that matches the approximate form that I have and when you go back and look at that chart I see 1/s - a and that is the Laplace transform of e^at, s - a^2 is the Laplace transform of te^at.1955

For higher powers this s - a^3, I figured out there is a generic Laplace transform of t^n t^at, in order to get S - a^3 here I have to take my n=2, I plug in n=2 there and I get 2 factorial is 2.1974

In order to match that 2 in the numerator, I had to factor this 4 into 2 x 2 and that is why instead of 4 here, we just get that 2 became 2t^2 e^at, but a was just 1.1993

That came in as my exponent in the e^at in then here I have 3t^at and 4e^at and putting all those terms together gave me the original function whose Laplace transform was this rational expression that we were given.2011

Let us try one more example here, we are going to find the inverse Laplace transform of 4s^2 + 4s/s +2 x s^2 + 4, our first step here is partial fractions and try to break this up into more manageable pieces.2030

I'm kind of relying on you to remember the partial fractions from calculus 2, got a lecture on partial fractions in the calculus 2 lectures here on www.educator.com, if you are still rusty on that.2052

Remember we want to set our generic partial fractions expansion form S + 2/s^2 + 4 and while we try to factor the denominator but we can not factor it any more than it is already been given to us.2063

We got an s^2 + 4 in the denominator that does not factor anymore, we just got to write this as a/ s + 2 and then we have a term over s^2 + 4 and we are going to have to call that BX + c.2086

Our goal is to find the a, b, and c as the goal of partial fractions and we are going to multiply both sides by our common denominator, in order to get rid of the denominators s + 2 x s^2 + 4.2107

When we do that, on the left hand side we get 4s^2 + 4s, on the right hand side we get a x s^2 + 4 + bx + c, I wrote x of course I was thinking in terms of calculus when we always use x as the variable.2125

In this case it is an s, b x s + c and we got s + 2 and now essentially we want to solve for a, b, and c, it is a little bit messy but I see something quick that I can do and find out one of my coefficients right away.2143

That is if I plug in S = -2, that will make all these terms dropout, I'm going to go ahead and do that, S = -2 and on the left side 4s^2 will be 16, it is 4 x 4 + 4 + s - if s is -2, minus 8 is equal to a x s^2 + 4 is 8.2165

The whole point here is that the (bs) + c terms get multiplied by zero and therefore they dropout and we get a x 8=8.2194

Right away we figure out the a is equal to 1 and that is as far as we can take it in terms of cleverness, now we have to actually expand out that expression on the right and it is a little messy but it is not too bad.2209

4s^2 + 4s, I already figured out that my a is one, I'm going to go ahead and write s^2 + 4, I'm going to expand out this (bs) + c x S + 2, that is (bs)^2 + 2(bs) + (cs) + 2c and I'm going to sort that out into powers of s's.2224

I see I got 1 + b x s^2 and takes care of that term and that term, now my S term are going to be 2b+ c is going to be my s term and my constant terms is going to be for 4 + 2c, this is still equal to 4s^2 + 4s.2254

And from my s^2 terms assuming that 4=1 + b , I think I'm going to skip to the constant terms because that looks a little simpler, constant terms tell me that on the left I have zero as a constants.2282

On the right I have 4 + 2c and that is probably enough that I can solve b and c, I would not worry about the S terms because I do not need it, but I see a 4= 1+ b, then b=3 and if 4 + 2c is equal to zero then my c= -2.2301

That means that I figured out all my coefficients a, b, and c and I can complete my partial fraction expansion 4s^2 + 4s/ s+ 2 x s^2 + 4 is equal to, it was a/s + 2 that is reading up here.2325

My a was one, 1/s +2+ (bs) + c, reading here might be as 3, 3s - 2, c is -2 and I got an s^2 + 4 here, so that is as far as I can go with the partial fractions.2351

I want to look back in my Laplace transform chart and see if I can find any functions that generally match the terms that I got here.2373

I will look at my Laplace transform chart, remember I posted that chart at the beginning of the lectures, we can flip back a few slides and you can see the chart posted back there and let us see what we have here.2385

Something that matches these terms that we have, I see the Laplace transform of e^at is 1/s - a, the Laplace transform I see for cos(bt) is S/s^2 + b^2, I picked that one because it looked like it might match the second term of our partial fraction expansion here. 2407

And Laplace transform of sin(bt) is equal to b/s^2 + b^2 and I'm just going to make a note here I see I got s^2 + 4 which means b is going to be 2. 2440

If b is equal to 2, then this will be equal to 2/s^2 + 4 and our cos(bt) would be just S/s^2 + 4, I think that is going to be enough to figure out my inverse Laplace transforms, this last term here, I'm going to separate it out and write this as 3s/s^2 + 4 -2/s^2 + 4.2462

I think I can write down my inverse Laplace transform f(t), I see 1/S +2 and I figured out the Laplace transform of e^at is 1/s - a, I guess my a there will be -2, this will be e^-2t, now I have 3s/s^2+ 4.2498

That is 3cos(bt), b is 2, cos(2t) -2/s^2 + 4, 2/s^2 + 4 is sin(bt), so minus sin(2t), it is not 2 sin of 2t because this 2, we do not want to think of it as a coefficient, that 2 becomes that 2 right there.2521

We do not have another 2 outside the sine, that is it, we got our inverse Laplace transform of the function that we started with, let me go back over that quickly and remind you each of the key steps there.2549

The first key step is remembering how to do partial fractions, remembering what you learn in calculus 2 about partial fractions, we want to expand this function out into its partial fraction expansion.2568

That is a/ s + 2 and since we can not factor s^2 + 4, its expansion is just (bs) + c, to figure out the a, b, and c we want to clear denominators, that is what I'm doing here is multiplying both sides by the common denominator.2580

That gives us on the right eight a x s^2 + 4, (bs) + c x s + 2 and I notice I'm really trying to solve for three variables but I can get one of them quickly by plugging an s=-2.2599

I noticed that by looking at that, if I plug in s=-2 that term will dropout, s=-2 on the left gives us 16-8, on the right made this one term drop out and it gives us a x 8 , quickly we figured out that a=1.2613

The other coefficients took more work, I took the a=1, I plug it back in up here and we got s^2 + 4 and then I expanded out (bs) + c x s + 2.2630

That is how I got all of these terms , I sorted them out into terms with s^2, terms with s, and terms with the constant and I equated like terms on both sides, on the left I first looked at s^2.2644

On the left I got 4, that 4 right there and on the right I got 1 + b, it told me quickly that b was 3, I skipped over s because it looked a little more complicated, I skipped right to the constants.2663

On the left I got zero, that is coming from that zero there, on the right I got 4+ 2s, that is where that came from and that was equation I can quickly solve for c being -2.2677

I took that a, b, and c and I plug them back into my original guess a, b, and c, that is how I figured out this partial fraction expansion, that is the end of the calculus 2 portion.2691

Then I looked back at my Laplace transform chart and the goal of that is to find some functions that match this expansion, when I looked at my Laplace transform chart, I saw 1/s -a coming from e^at and I also saw s/s^2 + b^2 coming from cos(bt).2702

b/s^2 + b^2 coming from sin(bt) that look pretty hopeful and I noticed since I had a 4 here, I must be using b=2 because it is always s^2 + b^2 and if I plug in b=2, for the sin I will get 2/s^2 + 4.2725

For the cosine I will get s/s^2 + 4 and in order to invoke those I separated this out into the 3s part and the 2, the 3s part gave me 3 cos(2t) and the 2 gave me sin(2t) not 2 sin(2t) because there was already a built in 2 here.2744

That 2, I used it to match that and I did not have another 2 left over as a coefficient, meanwhile this 1/s + 2, I see that I had a =-2 here, that just turns into e^-2t.2768

I put all those parts together and I get the original function of t for which if you took the Laplace transform you would have got this function of s that we started out with.2783

What I really found here was the inverse Laplace transform of that function of s and my answer is a function of t.2794

That is the end of this set of lectures on Laplace transforms and inverse Laplace transforms, in the next lecture here we are going to see how we can use Laplace transforms to solve differential equations and solve the initial value problems.2802

That is the next lecture in this series on differential equations here on www.educator.com. My name is Will Murray and thanks for watching.2818

Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and today we are going to be looking at Laplace transforms and using them to solve initial value problems.0000

Let me show you how that works out, the idea is that you will be given an initial value problem of this swarm (a)Y″ + (b)Y′ + (c)Y=g(t), then you will also have two initial values.0012

y(0) is equal to y0 and Y′ 0 is equal to y0′ we have two initial values and you have the actual differential equation and the idea is that you take the Laplace transform of both sides of the differential equation.0027

In particular we are just looking at this part right now, we are going to take the Laplace transform of both sides, we are going to take the Laplace transform of (a)Y″ + (b)Y′ +cy.0043

Remember that the Laplace transform is linear, that splits up into a times the Laplace transform of Y″, b times Laplace transform of Y′ and c times Laplace transform of y.0055

Over on the right hand side, we had g(t) in the original differential equation, we get the Laplace transform of g(t) on the right hand side.0069

Let me show you how we use that, the idea is that we know what L(Y″) is, this is something that you can work out if you just fill around the Laplace transform a couple times.0078

It turns out to be s^2 x L(y) - s x y(0) - Y′ (0) and then we have a similar expression for L(Y′), it turns into S x L (y) - y(0), remember the y(0) and Y′(0) those are all quantities that we know from the initial conditions.0092

For each of these we can plug in numbers that we get from the initial conditions, we write that down numbers from initial conditions, those will all be given to us in the original problem.0123

We are going to plug in L(Y″) L(Y′) in by using these two identities and will get an equation that we can solve for L(y), I will end up with L(y) is equal to blah- blah -- some kind of function that will be in terms of x.0143

Then we are going to take the inverse Laplace transform and start with this function of s and go backwards to find y which will be a function of t and that will be our complete solution to the initial value problem.0159

That is the way it is going to work, let me mention one step here that is very important taking the inverse Laplace transform, that is something that we learned about in the previous lecture here on www.educator.com.0180

That is the lecture immediately preceding this one in the differential equations lectures series, if you do not remember that, if you did not watch that lecture recently.0192

You really want to go back and work through that lecture before you do this lecture because you want to be very comfortable with the inverse Laplace transform before we start using it to solve the the initial value problems.0202

In particular the problems that I'm going to solve today are closely tied to the problems that we used in that previous lecture, every example that we study today was a problem that we already used in the previous lecture to study the inverse Laplace transform.0215

We are going to be using the answers from that previous lecture as part of our work today, if you have not looked at that previous lecture recently, you might want to go back and look at that.0231

In particular check out the answers to each of those problems because we are going to be using each one of those problems to solve our problems today on initial value problems.0240

Let us go ahead and see some examples of this, we are going to solve the following initial value problem Y″ + Y′ -2y=0 and then we have 2 initial conditions y(0) is equal to 7 and Y′(0) is equal to -2.0250

Let me remind you of the two identities that we had at the beginning of this lecture, we had L(Y″) this is something we learned at the beginning of the lecture is equal to s^2 L(y) - S x y(0) - Y′(0).0270

That was something from the beginning of lecture and we also had one that was relevant to L(y′) which was s x L(y) - y(0) what we are going to do with each of these is plug in the initial conditions that we are given.0293

The y(0) and Y′(0), for this first one we have s^2 L(y) - s x y(0) but y(0) is given to us to be 7, this is -7s - Y′(0) , Y′(0) -2, this is +2 and then over down here with L(y′) we have s times Laplace transform of y-y(0).0313

That is -7 and we are going to go to this differential equation and we are going to take the Laplace transform of both sides and what we get on the left is L(Y″).0344

I'm going to plug in what I have for L(Y″) that is s^2 L(y) - 7s + 2 + L(Y′) that is s x L(y) -7, this part all came from L(y″), this part all came from L(Y′) and -2y.0359

That is-2 times L(y) is equal to L(0), that is just zero, what we have here is an equation now and you want to think of the variable is being L(y), we are going to try to solve for L(y).0396

I'm going to factor out all the terms I can from L(y) I see I have s^2 times, let me see what other terms I have before I start writing down the L(y).0411

s^2, I see have s^2, I have a plus S and then I have a -2, that takes care of that term and that term and that term, all those multiplied by L(y).0424

I'm going to collect my other terms, I see I have a -7s +2 - 7 is -5 is equal to 0 and the goal here is to solve for L(y), I'm going to move the 7s -5 over to the other side.0443

I have s^2 + s -2 x L(y) is equal to now moving them over to the other side and I get +7s +5 and finally I can get L(y) is equal to 7s + 5/s^2 + s - 2.0462

What I need to do now is to do the inverse Laplace transform to figure out what the original y was, now this is exactly what we did in the answer to example 1 in the previous lecture.0490

I'm not going to go through the details of that because that was quite a bit of work back then we are going to take the inverse Laplace transform and this is exactly the one that we did for example 1 of the previous lecture.0505

I'm just going to give you a reference to that example 1 in the previous lecture, you can go back and check that out and see how the arithmetic worked out there and what you will find is that we derive y=3e^-2t + 4e^.0528

That is our solution to the original initial value problem that we we are given, 3e^-2t + 4e^t, let me go back and recap what we did there, what we started out by doing was taking the Laplace transform of both sides.0557

But then that give us an L(y″) on the left and we had an identity for that at the beginning of this lecture which was s^2 L(y) - minus sY(0) - Y′(0) and y(0) zero Y′ (0) those are the initial conditions given to us .0575

I plug in those numbers here to get -7s + 2 and we also had L(Y′) we had an identity for that at the beginning of lecture which is sL(y) - y(0).0595

Again I plug in y(0) given to me is the initial condition is -7, I took each of these two expressions, this one and this one, I plugged them in for Y′ and Y″ here and so I get s^2 L(y) - 7s + 2 , sL(Y -7)0607

This 2y gave us 2L(y) is equal to zero and what l want to do is think of L(y) as the variable and I want to solve for that variable.0632

I found all my terms that were multiplied by L(y) and that was s^2 + s -2, found all my other terms separated those out, move them over to the other side and then I divided by the coefficient of L(y) and I got an expression for L(y).0643

This point out have to take the inverse Laplace transform this is exactly what we did in the previous lecture, if we go back and look at example 1 from the previous lecture.0659

You will see the arithmetic for that where we did partial fractions on this expression that we looked back at our Laplace transform chart. 0668

What we end up with from the previous lecture is that y=3e^-2t + 4e^t, that is our answer for example 1, let us go ahead look at another one.0678

We had the initial value problem Y″ +3Y′ equal zero and then some initial values Y(0) is equal to -2 and Y′(0) is equal to three, we are going to start out just like we did before.0692

We are going to take the Laplace transform of both sides of the differential equation but we are going to need to know what L(Y′) and L(Y″) are. 0704

We have those identities from the beginning of the lecture L(Y″) is the same identity as before it would be the same every time. 0716

It is s^2 L(y) - s x y(0) - Y′(0), that is the same every time and now we fill in what the initial conditions tell us about y(0) and Y′(0), we got s^2 L(y) still nothing I can do about that yet, minus s x y(0) that is + 2 x y(0) term plus 2s - Y′(0) is - 3 there.0723

Then L(Y′) again using the identity from the beginning of this lecture is s times L(y) - y(0), I will fill in what I know about y(0); well that is -2.0761

Subtracting that will give me a + 2 and I'm going to plug both of those back into the differential equation, I will get s^2 L(y) + 2s -3, that came from my L(y)″.0785

Now +3 Y′, I'm going to multiply this by 3, +3s L(y) + 6, that came from calculating 3L(y)′ is equal to zero and remember you want to think of L(y) being the variable and we are going to try to solve for L(y) .0806

I'm going to final my terms of L(y), I see I have an s^2 + 3s x L(y) and now I'm just going to look at all the other terms, I see I have 2s - 3 and a +6, that is plus 2s + 3 is equal to zero.0831

I will move that over to the other side, minus 2s -3 over on the other side and now if I solve for L(y) that means I have to divide both sides by s^2 + 3s.0854

I get L(y) - 2s -3/s^2 + 3s, I need to take the inverse Laplace transform of this and we learn how to take inverse Laplace transforms in the previous lecture here on the differential equations lectures series on www.educator.com.0869

I will take the inverse Laplace transform and we did this one as one of the examples in the previous lecture this was example 2 in the previous lecture.0892

You can check back at the previous lecture and you will see the arithmetic that we went through to take the inverse Laplace transform for this example.0911

You will see what we figured out is that y itself is equal to -1 - e^-3t, that is our solution to the initial value problem.0931

Let me go back over that and just remind you what the steps were, we took the Laplace transform of both sides L(y)″ + 3(Y′).0948

We use the identity that we learned in the beginning of this lecture L(Y ″) is equal to s^2 L(y) - sY(0) - Y′(0). that identity always holds it very safe, you are going to use that same identity in every problem.0957

What you do with that is you plug in the initial conditions that you are given for y(0) and Y′(0), that gave me s^2 L(y), y(0) is -2, we got a plus 2s here and then Y′(0) is 3, we got -3 there.0972

L(y) we are going to use that because of this L(Y′) is equal to sL(y) - y(0), again that is the identity from the beginning of this lecture and what we can fill in is that y(0) is 2.0991

We plug both those back into the differential equation here and here, the L(Y′) gets multiplied by 3, that is why we got that 3 and the 2 turn into a 6 and you want to think of L(y) being the variable.1008

We segregate all the terms that have an L(y) in them and I see that I have an s^2 L(y) and 3sL(y) and then all the extra terms were 2s - 3, that - 3 in that 6 combined to give me a +3 here.1022

I wanted to move the 2s +3 over to the other side because I was trying to solve for the L(y), I moved those over the other side, they turn into negatives and I got L(y) equal to minus -2s -3/s^2 + 3s.1039

Now it is an inverse Laplace transform problem, in fact it is the exact one that we solved in example 2 on the previous lecture, go back and check that out if you have not seen that for a while.1057

What we figured out is that, that corresponds to the original function y(-1) - e^-3t, that is our answer to the initial value problem.1067

In example 3 we got another initial value problem Y″ + 4Y′ + 5y=0 and a couple of initial values y(0)=1 and Y′(0)=0.1080

We are going to work it out exactly the same way as the first two examples using the same identities that we learned back in the beginning of lecture.1093

L(Y″) is equal to s^2 L(y) - S x y(0) - Y′(0) and again I'm going to fill in the initial conditions for y(0) and Y′(0), this is s^2 L(y), now s x y(0), y(0) is 1, this is just -s.1101

Y′(0) is 0, there is just a zero for that term there, that will just drop right out, next we have L(Y′) and again our identity from the beginning of lecture this never changes, it is always the same s x L(y) - y(0).1135

In this case, if that is just s x L(y) and y(0) we are given that as an initial condition that is 1.1158

Now I'm going to take each of these and plug them into the differential equation, that means I have to multiply L(Y′) by 4, I get s^2 L(y) - s, because I already simplified it down to s^2 L(y) - s and now I have 4 Y′.1169

I'm going to put + 4s L(y) - 4, I multiply everything here by 4 to get that term right there, +5y, +5 times the Laplace transform of y is equal to 0 and remember L(y) you want to think of that as your variable, you want to solve for that variable.1195

Let us see what terms I have multiplied by L(y), I on the left I have s^2 L(y), I see a 4s L(y) and I see a 5 L(y) and let me see what other terms I have.1225

I see just -s and -4, -s -4 is equal to zero and if I move that over to the other side, on the other side I will get an s + 4 and my L(y) if I divide both sides by the coefficient there would s + 4/s^2 + 4s + 5.1241

That is my Laplace transform of the y that I'm looking for and to solve that out I need to take the inverse Laplace transform.1266

I synched up all these examples to the examples in the previous lecture on inverse Laplace transforms, this was exactly the function that we used in example 3 of the previous lecture on inverse Laplace transforms.1285

In example 3, in previous lecture which is where we solve this one out and what you will see is we figured out that the inverse Laplace transform of s + 4/s^2 + 4s + 5 is equal to e^-2t x cos(t) + 2e^-2t x sin(t).1301

That is our answer to the initial value problem, let me go over the steps there.1336

We started out with Y″ + 4Y′ + 5y=0, we want to take the Laplace transform of both sides there, that is why we needed to know the Laplace transform of Y″ and Y′.1349

We have identities for those, s^2 L(y) - sY(0) - Y′(0) and then we plug in our initial conditions y(0) is 1, Y′(0) is 0, our L(Y′) simplifies down to s^2 L(y) minus s and L(Y ′).1364

The identity from the beginning of the lecture is sL(y) - y(0), that is a universal rule and then we plug in y(0) is 1, that is where we got that term.1384

We plug all those back into the differential equations, we get s^2 L(y) - s. that is our L(Y′), this term right here is L(Y″) and then we had 4 L(Y′), that 4 became those 4's right here.1396

We had to multiply everything here by 4, such when we got the for SL of why -4 and then +5y that became + 5L(y) is equal to zero, you want to think of L(y) as your variable.1415

You want to separate out all the terms multiplied by L(y) and then collect the other terms -S -4, move those other terms over to the other side that is what we did here and then divide by s^2 + 4s + 5.1429

That became a denominator and now we get L(y) is a function of s and that is something that we need to apply the inverse Laplace transform to and this particular function is the one we studied in example 5 of the previous lecture.1445

If you go back and look at example 5 of the previous lecture, we started out with this function of s and we did a lot of arithmetic on it, some partial fractions in completing a square.1462

What we ended up with in the previous lecture was this function of ye^-2t cos(t) + 2e^-2t sin(t), what that tells us is that this is the solution to our initial value problem.1473

For example 4, we got the following initial value problem y″ - 2Y′ + y=4e^t and then a couple of initial conditions, y(0)=4 and Y′(0)=1.1494

This ones going to be very much similar to the other examples, we are going to take the Laplace transform of both sides, the one difference here is that we have an inhomogeneous differential equation.1508

Inhomogeneous means that that right hand side is not zero, we are going to have to take the Laplace transform of that as well because it is not just coming out of the zero.1518

But still the arithmetic kind of work out quite similar to the previous equations, let us see how that goes, we will need L(Y′), Y″ and we had an identity for that at the beginning of this lecture which was s^2 L(y) - sy(0)- Y′(0).1527

We have numbers for y(0) and Y′(0), I will go ahead and fill those in, s^2 L(y) - sy(0), that is 4s - Y′(0) is -1 and then we also have an identity for L (Y′).1554

Again from the beginning lecture these the two identities that never change s x L (y) - y(0) that is the identity from beginning of lecture that never changes is equal to sL(y) and my y(0) we are given that as our initial condition that was 4. 1573

I'm going to plug those back into the differential equation taking the Laplace transform of both sides, I see L(Y″), I'm going to g s^2 - L(y) - 4s - 1.1599

Now -2(Y′) that is minus 2L(Y′) is SL(y) - 4 + y, plus L(y) is equal to 4e^t , this is the Laplace transform of 4e^t.1614

If you go back and look at the first lecture on Laplace transforms, we figured out what the Laplace transform of e^t is and it is 1/s-1.1640

Basically the Laplace transform of e^at, go back and look at the first lecture on Laplace transform, I think it was two lectures ago here in the differential equations lectures series on www.educator.com.1652

It is 1/s - a, person, in this case we have a 4 as a coefficient, I'm going to write that as 4/s -1 and now just like the other problems, we are going to think of L(y) as our variable, we are going to try to solve this equation for L(y).1666

It is going to involve collecting some algebra, I have some terms for L(y) and I also have some constant terms, let me see what I have for L(y), I see I have an s^2 here, I see I have a -2s here, and here is an L(y), that is +1.1683

Let me collect the other constant terms, I see -4s, that is the only term of s I see, -4s, -1 and this -2 x -4 would give me a +8, 8-1 is 7.1708

That is combining a couple constant terms there and this is still equal to 4/S -1 and let me move these terms over the other side, the 4s + 7, if I move those over to the other side, on the left I will get s^2 -2s + 1 x L(y)=4/s-1 + 4s - 7.1728

I like to combine all if that over a common denominator, I'm going to take that 4s -7 and multiply by S -1 / S -1 and if I expand out that multiplication the 4s -7 x S -1, that is 4s^2 -7s - 4s, that is -11s.1762

Now I have -7 x -1 is + 7, I combine these terms together, I saw I have an S -7 in the denominator that was a whole point of combining them and I get 4 + 4s^2 -11s + 7.1787

On the left hand side, I got an L(y) and I think I'm going to factor that s^2 - 2s + 1, I know that factors as S -1^2 and if I keep working on the right I can combine this 4 and this 7 into 11.1812

I got 4s^2 -11s + 11 and on the bottom I have S -1 but I could also divide by S -1^2 combine those two together and I get S -1^3 for my L(y) that is just a pile of algebra, all with the goal of solving for L(y) into a function of s.1822

The point of this is that I can now take the inverse Laplace transform, take the inverse Laplace transform and remember that all the examples in this lectures are synced up to the examples from the previous lecture.1857

I worked out all the inverse Laplace transforms in the previous lecture, you can go back and check those out if you have not seen it recently but this was example 4 in the previous lecture where we worked out the inverse Laplace transform of this function.1883

It is 4s^2 -11s +11/ s -1^3, it was messy that is why I do not want to do the work again but we figured out that the original y as a function of t is 4e^t -3et^t + 2t^2 e^t.1908

That was some work in the previous lecture to calculate that out but it worked and if it does not make sense to anymore, maybe go back and look the previous lecture and work through example 4 and you will see how we got from this step to this step.1932

Let me remind you of each of the steps there, in case anything is still a little fuzzy, we wanted to take the Laplace transform of both sides, we need the Laplace transform of Y″ and we have an identity for that.1949

That is a universal identity that we learned that at the beginning of lecture and it is the same in every problem, s^2 L(y) - sy(0) - Y′(0) and then what changes in each problem is the values of y(0) and Y′(0).1963

We took the 4 for y(0), plug that in and we took the 1 for Y′(0) and plug that in, and then we have an L(Y′), that by our identity from the beginning of lecture is SL(y) - y(0).1979

We plug in y(0)=4 and then we plug those back into the original differential equation, there is my L(Y″) and there is my L(Y′) just copying that from above but of course with a coefficient of -2.1994

Now from the differential equation there is my L(y) but since the right hand side is not zero, we have to take the Laplace transform of the right hand side as well, that is L(4e^t).2018

And something we learned in the first lecture on Laplace transforms is the L(e^at) is 1/s -a, if you are little rusty on where that comes from, check out the first lecture on Laplace transforms it is still in this differential equations lecture series.2030

It is a couple of lectures ago and you will see where that, I forget my s here in transforms.2050

You will see where that L(4e^t) turns out to be 4/ S -1, now we have a slightly nasty algebra problem, what we want do is to solve for L(y), you think of L(y) as being the variable.2067

That is why I collected all my terms multiplied by L(y), collected all the extra terms, that 7 by the way came from this -1 and this -2 times this -4, -1+ 8.2082

That is equal to 4/s - 1 and I move this 4s -7 over to the other side here, put it over a common denominator which meant I had to multiply it by S -1, that is me expanding out that multiplication right there.2096

When you combine it, the 4 and the 7, that 4 and that 7 combined and gave you that 11, meantime on the left hand side I have factored S^2 - 2s + 1and s-1^2 and then divide both sides by S -1^2.2114

That is why you end up getting S -1^3 on the denominator on the right and now we have to take the inverse Laplace transform which is another batch of algebra which we worked out in the previous lecture.2130

The previous lecture example 4, we start with exactly this function and then we did partial fractions on this function, expanded out into the inverse Laplace transform and we converted it into 4e^t - 3te^t + 2t^2 e^t.2142

That is our answer to the initial value problem, on our last example here, we got the following initial value problem Y″ + 4y=ae^-2t and then we have a couple initial values y(0)=4, Y′(0)=-4.2160

We are going to take the Laplace transform of both sides there, on the left we are going to need to know L(Y″), we have an identity for that from the beginning of this particular lecture.2183

That was s^2 L(y) - s x y(0) -Y′(0) and we can fill in the numbers that we have here, this is s^2 L(y) and we cannot fill in anything for that yet, but s x y(0), y(0) is 4t, that is -4s and -Y′(0).2197

Y′(0) is -4, that is + 4, it looks like we are not going to need the Laplace transform of Y′, I'm going to go ahead and plug this right into the differential equation.2223

We get s^2 L(y) -4s + 4 + 4L(y), that term right there is coming from here because if we took Laplace transform both sides and then on the right, we have to take the Laplace transform of 8e^-2t.2234

Let me remind you of the Laplace transforms we did back 2 lectures ago here in the differential equation series, in the Laplace transform lecture you will see that in the list of lectures over there.2259

In the Laplace transform lecture we figured out that the Laplace transform of e^at is 1/s - a, check that out in the original Laplace transform lecture if you do not remember where that comes from.2276

That is just a couple lectures ago here on the www.educator.com and e^-2t our a is -3 a is -2, that is 1/s + 2 and we still have an 8 here coming along as the coefficient.2297

We really got an algebraic problem we are trying to solve for L(y), think of that as being the variable and solve for that, I'm going to collect my terms there multiplied by L(y), I see we have an w^2 + 4 x L(y) - 4s + 4=8/s + 2.2319

I'm trying to solve for L(y), I'm going to take the -4s + 4 and move it over to the other side, that will turn into + 4s - 4, in order to put that over a common denominator I have to multiply that by S + 2 / S + 2.2345

I need to expand that out, this 4s -4 and the s + 2, if I expand that out and I will get 4s^2 - 4s + 8s, that is a + 4s and then -4 times +2 is -8, if I combine that all together, I have a -8 and a +8 here, those will cancel each other out.2366

I will just get for 4s^2 + 4s/s + 2 and that is still equal to s^2 + 4 x L(y) and remember we are solving for L(y), I'm going to divide by the coefficient of L(y), s^2 + 4.2393

L(y)=4s^2 + 4s/ s^2 + 4 x s + 2.2413

That is the end of that step of algebra, the next step is now that I know what the Laplace transform of y is in terms of s, I need to take the inverse Laplace transform.2431

I synced these examples up to the examples in the previous lecture, this was example 5 in the previous lecture, see example 5 in the previous lecture which was called inverse Laplace transforms.2452

What we did was we started with this function of s and we did partial fractions on it, we worked backwards and we finally found out a function for y in terms of t,e^-2t + 3 cos(2t) - sin(2t).2475

That is really exploiting a result from the previous lecture, example 5 from the previous lecture if you want to look it up and see where that came from, see how we did that.2497

We worked out all the arithmetic back there, it all worked out do not worry that is really the end of that problem, let me show you the steps we went through to get there.2507

We took the Laplace transform of both sides which means we had figure out L(Y″), we had this identity from the beginning of lecture which always works s^2 L(y) - sy(0) - Y′(0).2516

We fill in our values for y(0) and Y′(0) which is 4 for y(0) and -4 for Y′(0) of course the -4 turns into a +4 because that negative sign cancel that negative sign and gave me a positive there.2531

We plugged that back into the differential equation s^2 L(y) -4s + 4 and then that 4y gave me that term right there.2549

Then we also have this right hand side, we had to take the Laplace transform of 8e^-2t and we were called from the original Laplace transform lecture, not the previous one but the one before that.2560

We first started talking about Laplace transforms, we figured out the Laplace transform of e^at is 1/s -a, in this case our a is -2 and we get 1/s +2 and then there is also an 8 there.2572

We just bring that along too and what we find here is kind of a big algebraic expression, you want to think L(y) as the variable and solve for that, just kind of sort out all the s's and just get a nice expression for L(y) in terms of s.2587

That is what we are doing here, we moved the -4s + 4 over to the other side, that is where that came from, I want to put it over a common denominator, I saw and s + 2 here, I multiply top and bottom here by s + 2.2603

That is a little messy I had to expand 4s - 4 x s + 2, that expanded out into 4s^2 + 4s -8 and when I combined that with this 8, the 2 8 cancel each other out which is why there is no constant over here.2621

We just get the 4s^2 + 4s still over S + 2, in the meantime I still have s^2 + 4 x L(y) on the left and when I divide that across that joins the denominator here and we get L(y) is 4s^2 + 4s/s + 2 x s^2 + 4.2639

We had to take the inverse Laplace transform which will be another really healthy dose of algebra here but it is exactly the example 5 in the previous lectures, go back and check that out, work that through.2660

What we figured out is the inverse Laplace transform of that is exactly this e^-2t +3 cos 2t - sin (2t), I see that I wrote that is if it were not exponents.2674

Let me write that a little nicer down here, 3cos(2t) - sin(2t) is our final answer there.2688

That is the end of our lecture on using Laplace transforms to solve initial value problems and that is actually the end of this chapter on Laplace transforms, I really appreciate you watching.2698

My name is Will Murray. You are watching the differential equations lectures series here on www.educator.com, thanks for joining us.2709

Hi and welcome back to the differential equations lectures here on www.educator.com, my name is Will Murray and we are going to be studying systems of linear differential equations.0000

Before we jump into that, I think it would be good if we did a quick lecture on review of linear algebra because you really do have to remember your linear algebra concepts in order to solve these systems later on.0010

In this lecture we are not really going to be talking about differential equations per se but we are just going to review some linear algebra concepts especially eigenvalues and eigenvectors.0022

You will be using those to solve differential equations to subsystems of differential equations later on, let us jump right in here. 0031

Let me remind you first of all how you multiply matrices, I'm not really going to teach this to you from scratch, I assume you done a little bit of linear algebra before but this will be just a quick review of it is been a while since you done it.0042

The idea here is we will start within by m x n matrix, what that means is m rows and n columns, this would be m and then the number of columns here is n and you can multiply that by an m x p matrix.0054

Here the number of rows is n, n rows and p will be the number of columns and your answer will be in m x p matrix, you will get m rows and p columns, the way you want to think about that is you can multiply (m x n) x( n x p).0074

The middle two numbers always have to match this n always has to match, and then in the answers those cancel each other away and you always get an m x p matrix.0099

Maybe that is easier to think about if we actually put some numbers here, what I got with this example is this first matrix is a 2 x 4 matrix, the m is 2 rows and n is 4, 4columns. The second matrix is 4 x 3, n is 4 rows, p is 3 columns.0111

When we multiply a 2 x 4 matrix with a 4 x 2 matrix, a 4 x 3 matrix, that is legit because the 4's matchup, you always have to have those inner two numbers match up.0132

You multiply 2 x 4 x 4 x 3 matrix and you think of those 4's were cancelling each other out and the answer will be a 2 x 3 matrix, the m here is 2 rows and the p here is 3 columns, that is how you check out the sizes.0145

The way you actually multiply them is you, to find an entry in the answer, you look at the corresponding row of the first matrix and the corresponding column of the second matrix.0166

Here to find this entry, if this row were A, B, C, D and maybe then this row is E, F, G, H and this column were X, Y, Z, W, the way you get that entry first row times first column is you multiply A, B, C, D times X, Y, Z, W.0184

This would be AX + PY + CZ +DW and you go through and you find each entry in the answer matrix the same way, for example the way you find this entry well that is the second row.0210

I look at the second row and that is the second column, I look to the second column of the second matrix, second row of the of the first matrix, second column of the second matrix, maybe that is R, S, T, U.0229

I'm multiplying E, F, G, H times R, S, T, U, to get that entry I would do ER + FS + GT + HU, I'm getting that from this row and this column and you go through and you find each entry in the answer matrix that way.0248

We will do an example, if you are a little rusty on this, I think it will make a lot more sense when we do the example on actual numbers, let me go ahead and remind you some other things we are going to be studying.0274

You can take determinants as follows, the determinant of a 2 x 2 matrix is really easy, you just cross multiply positive in that direction and cross multiply negative in that direction, you will get A x D down the main diagonal - B x C down the back diagonal.0286

That is the determinant of a 2 x 2 matrix, determinant of a 3 x 3 matrix kind works the same way but it is a lot more complicated, again you try to go down the diagonals here, A, E, I and then you want to go B, F, we are running off the matrix.0304

What you do is you copy the first two columns on the other side, think of it as wrapping around to the other side, that first column is A, D, G, I will copy that over here, ADG and that second column is B, E, H.0323

Let me switch colors here so that you will be able to see what I'm doing, I got A, E, I positive, B, F, G positive and C, D, H positive, that is why I got the AEI, BFG, CDH positive from those three positive diagonals.0342

Then I go down the negative diagonals CEG that is a negative, AFH that is a negative, and BDI that is a negative, if you look at whether were negative terms here that is AFH that comes from this one right here.0365

The BDI comes from this one right here, that is a negative and CEG, it is a little hard to read now but it is coming from these terms right here, that is a negative.0385

It is a little complicated to take a 3 x 3 determinant and there are other ways to do it, expansion all on row or column, you might have learned that in your linear algebra class and that is legitimate.0398

For me the easiest way is just to remember those three main diagonals and three back diagonals, main diagonals are all positive, back diagonals were all negative.0409

The next concepts that we need to remind you about for linear algebra is eigenvalues and eigenvectors, the idea here is that A would be a square matrix, let me draw a little picture of this A would be a square matrix.0419

V will be a column vector an n x 1 column vector, that is vector b and r is just a number, r is just a scaler and then V is still the same column vector.0436

If you multiply an n x n matrix times an n x 1 vector, remember how those middle numbers cancel and you get an n x 1 vector, if A x V is the same as what you would have gotten just by multiplying V by a scaler.0450

In that special situation that is the definition of an eigenvector, V is the eigenvector and R is the corresponding eigenvalue, that is what an eigenvector and eigenvalue mean.0468

It is just means that A times that vector is equal to a scaler times that vector, this definition of eigenvectors does not really show you how to find them, let me go ahead and show you how to find an eigenvalue and eigenvector.0481

The way do it is you look at the equation A - r(i)=0, remember (i) is the identity matrix, that means you look at the matrix A and you subtract an R down the main diagonal.0496

You subtract off R's down the main diagonal and then you take the determinant of that matrix and set it equal to 0, what you will get is what is called the characteristic polynomial in R.0513

This characteristic polynomial will just be a polynomial equation the you will solve and you will get one or more values for R, you will get a couple values for R and those values are the eigenvalues of the matrix.0527

For each one of those eigenvalues you got a find the associated eigenvector, you take that R, you plug it back into A - R(i) and you try to solve the generic vector A - R(i) times V is equal to the 0 vector.0543

Then you try to find those corresponding eigenvectors V, that is the way that works but I think it will make a lot more sense if we do some examples of eigenvalues and eigenvectors.0560

One thing to notice, I will mention it here although we will see an example of it later is we might find more than one eigenvector, we might find multiple eigenvectors and we will see an example of this.0573

You will see what that looks like, you might find multiple eigenvectors corresponding to a single eigenvalue, you will see what that looks like as we do some examples later on.0591

I think it is time we jumped into some examples, first example we are just going to practice multiplying matrices, we got this matrix has 2 rows and 4 columns, that is a 2 x 4 matrix, two rows and four columns, 0616

This other matrix has 4 rows and 3 columns, that is a 4 x 3 matrix, that means it is legitimate to multiply these 2 matrices, a 2 x 4 times of 4 x 3, you have to think of the 4's as cancelling, that is going to give us a 2 x 3 matrix.0641

I'm going to set up a 2 x 3 matrix as answer here, there is going to be two rows and three columns in my answer matrix and to get each entry, what I do is I multiply a row of the first matrix by column of the second matrix.0660

For that first entry, the first row, first column, I have 1, 2, 0, I give myself a little more space here and then times 2, 2, 3, 0 and I'm going to add those up.0688

1× 2 is 2, 2× 2 is 4, 2 + 4 is 6 and if I add those up I get a 6, that is why I put a 6 in that first entry there.0717

In that next entry, I got still the first row, the second column, first row time second column again 1, 2, 0, -1 times the second column is 0, 7, 1 and -4 and then I add them up.0729

I see I got 0, 4, 14, 0 and 4, that is 18 and means that second entry is in 18, for the last entry on the first row I look at the first row of the first matrix times the third column of the second matrix.0758

It is always rows of the first matrix times columns of the second matrix, 1, 2, 0, -1 times 5, -3, × 2 and 1 and add them up and I get 5 - 6 -1, that is a negative, no -1-1 so that is -2, 6, 18, -2 in that entry .0775

To find the second row, I take the second row of the first matrix and then I multiply that by each column of the second matrix, 3, -2, 0, 4 times 2, 2, 3, 0, add those up and I get the 6, -4, I get 2 for that first entry.0810

Now bottom column, middle row, 3, -2, 0, 4, 0, 7, 1, -4, add those up I see I got 0, -14, 0, -16, -14 + -16=-30.0845

Finally bottom row and the right hand column is 5, -3, 2, 1 add those up and I see I got 15, + 6, + 4, 15 + 6=21 + 4=25, I fill that one in there and that is my product matrix 6, 18, -2 and 2, -30, 25, that is my answer. 0879

Let me remind you how we did that, we start out with a 2 x 4 matrix, 2 rows and 4 columns multiply it by a 4 x 3 matrix, in order to be able to multiply them, those numbers in the middle have to match up.0923

Essentially that is saying that the length of these rows is equal to the length of these columns and they have to be equal so that you can match them up term by term.0936

Those 4's match up and in the answer matrix you drop the 4's out and you make it 2 x 3, that is why I set this matrix two rows and three columns and to get each entry in the answer matrix.0944

You look at the corresponding row of the first matrix multiply by the corresponding column of the second matrix.0959

For example this 18, that is the first row second column, I go first row that is this one right here 1, 2, 0 -1 and second column 0, 7, 1, -4, I did 1, 2, 0 -1 x 0, 7, 1, -4 multiply them together and add them up that is where that 18 comes from.0967

In each other entry comes the same way where you look at the corresponding row of the first matrix, column of the second matrix, multiply them together and get your answer.0996

The second example we got a system of differential equations, we are not actually trying to solve it but we are trying to write it as a system of equations, we are trying to write it in matrix multiplication form.1011

Let me show you how we can work that out, if you look at this middle column for example or middle row 3x + 2i + 12z + W, 3x + 2y + 12z + + w.1024

Let me show you how you can write that in matrix multiplication form as the row 3, 2, 12, 1 multiplied by the column x, y, z, w because if we multiply that out term by term we get 3X +12 + 2y + 12z + w.1043

What we are going to do is make all the coefficients of each of these equations rows of a matrix, we are going to multiply it by the column XYZW, let me write that out.1066

I see in this first row here, I got an x, the coefficient is 1, I will think of that as 1x + 0y + 2z + 0w, 1, 0, 2, 0 and in the the second row I got 3, 2, 12, 1 reading the coefficients, third row is 1, 1, 5, 1, take that matrix and I multiply by the column XYZW.1080

It might look like the sizes do not matchup there because the column is longer than the coefficient matrix but let me show you how they do, this coefficient matrix is 3 rows 4 columns that is a 3 x 4 matrix.1119

This column vector is 4 rows, 1 column, that is a 4 x 1 matrix and the important thing to multiply matrices is that those internal numbers match up, that 4 matches up with a 4, that 4 does really does match up.1135

We can multiply a 3 x 4 x 4 x 1 because those inside numbers match and our answer matrix is going to be the outside numbers 3 x 1 which means it is the three rows one column.1152

A single column with three rows and I can see on the right here that I got my answers are 5, 16, 5, what I have done here is I have given you a matrix form for that system of equations.1166

If you expanded out this matrix multiplication then it would turn into nothing more nothing less than these three equations here, let me remind you how that happened.1184

For each of these equations I want to write it as a row times a column, what I was doing right here was illustrating with the second equation 3X + 2y + 12Z + w, we can write that as the row 3, 2, 12, 1 times the column XYZW.1199

If we do that for every row, think of this first row as having a 0y and 0w, the coefficients are 1, 0, 2, 0, second row 3,2 ,12, 1, third row 1, 1, 5, 1 multiplied by the column X,Y, Z, W.1218

We got a 3 x 4 matrix that can multiply by 4 x 1 matrix, think of the 4's cancelling and your answer will be a 3 x 1 matrix and those are the answers that we get over on the right hand side, the 5, 16, 5.1235

In example 3, we are going to do something kind is similar to the previous example where we are going to write a system of differential equations in matrix multiplication form, here is the system we are given, x1′=2x1(t) - x2(t).1253

A similar equation for x2′, the first thing I'm going to do is just not write the t's at least the parentheses t's, I'm going to write this as x1′=2x1 - x2 + e^t.1269

I'm not changing anything there, I'm not cancelling the t's, we are going to remember that those t's are still in there, I'm just trying to write it a little more simply.1285

Then x2′=2x1 + x2 + sin(t) and I'm going to focus on this part right here because I can write that as a matrix multiplication, the same way we did in the previous example, in example 2.1293

You might want to go back and check on example 2 if you do not really remember how that one worked out, but I can write this as matrix multiplication if I put a column x1 x2, then my coefficients are 2, -1, 2 and 1.1314

I'm looking over here at the coefficients 2, -1, 2, 1 and if I multiply out that matrix multiplication, I would see I had 2, -1 times x1 x2, that would give me 2x1 - x2.1334

I also want to add on a column for the e^t and sin(t), on the left I see I have x1′, x2′, now we are going to introduce a little notation here, we use bold x, x with a bar over it to mean the column vector x1 x2.1350

If we do that then we will call X′ would just be the derivatives of each of those, x1′ x2′ and that lets me rewrite this equation on the left I have X vector X′ is equal to 2, -1, 2, 1 times vector X.1378

Because we have x1 x2 + e^t and sin(t) and from there I have a nice matrix multiplication form of my differential equations, let me remind you how we arrived at that.1399

First we suppressed each of these t's, it is not that they are not there anymore, we are just not going to write them anymore, we are not cancelling them, we are not going to write them anymore.1426

We got x1′ is equal to 2x1 minus x2 + e^t, x2′ is 2x1 + x2 + sin(t) and then we are going to write this in vector form where we write these coefficients as a matrix.1437

This is just like the previous example, example 2, if this step looks a little strange to you, sorting this out into matrix form, you might want go back and look at example 2 again, it might make this make a little more sense.1452

We wrote x1 x2 as a vector, wrote the coefficients 2, -1 and 2, 1, separate out the e^t and sin(t) as its own vector and then we have X1′ and x2′, we are going to call X vector x a shorthand for x1 x2.1463

Vector X a shorthand for x1′ x2′, that means on the left we have vector X′, still that same coefficient matrix 2, -1 and 2,1 and then vector x e^t sin(t) .1482

That is the vector form for this system of differential equations, what we are going to do in the next set of lectures here on www.educator.com is we are going to start solving these things.1497

We are going to figure out how to solve this system of differential equations using eigenvalues and eigenvectors.1507

We are not quite there yet, if you are excited to jump forward and figure out how to solve these, you might want to skip to the next lecture which is on how to solve systems of differential equations.1514

What we are going to do now is do a little more practice finding eigenvalues and eigenvectors and just make sure that we are up to speed on that.1523

In example 4, we are going to find the eigenvalues and eigenvectors of the matrix 2, 3, -1 and 6, and remember the way do that is you look at the determinant of A - R(i) and you set it equal to 0.1533

Of course R(i), (i) is just the matrix 1, 0, 0, 1 that is the identity matrix and R(i) is just R 0, 0, R, effectively what you are doing here is you are subtracting R's off the main diagonals.1551

A- R(i) is equal to 2-R(3) - 1 6 - R, we are just subtracting R's off the main diagonals and the determinant of that, you cross multiply, let me do that in another color so it does not get too confusing.1568

We cross multiply positive in the forward direction, negative in the backwards direction and we get 2- R x 6 - R - 3, it is +3, and we set that equal to 0.1592

I'm going to expand that out, I see I'm going to have a +R^2 -6r -2r, that is -8r and then +12 + 6 × 2+ there is another 3 there,+15 is equal to zero. 1611

If I factor that I can see the that factors into r -3 and r -5, my eigenvalues there are 3 and 5, those are both eigenvalues of this matrix, this is common by the way, you expect a matrix to have more than one eigenvalues in general.1629

Let us see how we find the corresponding eigenvectors, what you do is you plug A- R(i) back into the matrix for each value of R and then you solve for A- R(i)V is equal to zero.1650

When r=3, A- R(i) we will do 2-3, 3 -1, 6-3 and that gives us -1, 3, -1, 6 - 3 is 3 and we are going to try to solve A- R(i) x V is equal to 0. 1667

If you remember from your linear algebra class, what you have to do now is to row reduce the matrix, I can see right away that this bottom rows can go to zero because it is the exact same as the top row.1693

I have zeros in the bottom row, I can subtract top row from the bottom one, maybe I will multiply the top row by -1 to make a little more cleaner, 1 -3 and since I'm trying to solve A- R(i) x V=0.1708

I'm going to write that as x x y =0 and if I expand that out I see I got x -3y is equal to 0, the way solve this in your linear algebra class is through the use of free parameters, I'm going to say y=t.1721

I got x - 3t=0, t is a free parameter, X is equal to 3t and xy as a vector is equal to t and 3t, I mixed those up because the x is 3t and y was t and we could factor out t and we just get a 3(1) there.1743

This will work for any value of t, any value of t will give us an eigenvector, but to make it as easy as possible, let us just take T is equal to one, my eigenvector for r=3 is 3, 1.1774

I'm going to do the same thing for R=5, again I'm going to plug in A- R(i), I'm going to subtract 5 off the main diagonal from the original matrix 2 -5 is -3, 3, -1, 6 - 5 is 1.1793

I'm going to do row reduction on this matrix, think the easiest things to switch the rows and multiply the bottom row by -1 as I switch, that will give me 1 and -1, you notice that the top row was just a multiple of the bottom row.1814

I'm going to drop that right out, I'm doing several steps of row reduction in my head here, if you are little rusty on row reduction, you might want go back and check the www.educator.com lectures on linear algebra.1829

You will see lots of practice on row reduction, I'm doing this a little fast here and I'm also going to multiply this times XY and get zero and that tells me the x - y is equal to zero.1840

If I use a free parameter y=t, X would have to equal y, X would be t as well and XY would be t times 1, 1 and again we can just take T to be 1, our xy would be, our eigenvector would be 1,.1857

Let me summarize here, the eigenvalues are 3 and 5, each one has a corresponding eigenvector, the vector for 3 was 3, 1 and the vector for 5 was 1, 1.1879

Let me summarize this and teach you a little trick that sometimes can be helpful here, what we did here was we started out by working out the determinant of A- R(i) which means you subtract an R off the main diagonal.1912

That is what we did here, that is what this - R and that -R came from, take the determinant means cross multiply, you will get 2-r x 6-r -3, that negative and that negative cancel each other out and gave us a +3 there.1928

Expanded that out, we got a polynomial which factored pretty easily into 3 and 5, and then for each one of those we plug those back in as the -R and we got matrix and then we reduce that matrix using our row operations for linear algebra.1945

You might want to review your row operations for linear algebra if you are having trouble reducing a matrix and then we used free parameters to set up a system of equations y=t, x would have to be 3t and then we get xy is t x 3(1).1964

We can just take t=1 to be 3 x 1, now let me show you a little bit of a shortcut you could have done there, what we are looking for is a vector XY, when we multiply 1 A-3 times it will get zero.1983

Sometimes what you can do is a shortcut here is you just put 1 in for the bottom component y, and then you see what the top component would have to be in order to make it come out to be zero when you multiply it by this vector.2002

Here I'm looking at one times something -3 8 1=0 and I see that it is something would have to be a 3 and my eigenvector is 3,1.2019

That is a little trick the you can sometimes employ, it means you put one in for the bottom component and then you just figure out what the top component would have to be in order for it to come out to be zero when you multiply it by this row here.2032

Let us try that out with r=5, if we look at A- R(i) subtracting R off the main diagonal we get -3 here and 1 here, the other entries are the same and that reduces our row operations to 1, -1.2048

If we just take that y entry to be 1 and I think what would that X entry have to be in order to get 0 when we multiply this row by it, I see that if that x were 1, then 1× 1 - 1× 1 would give us 0.2064

That would be a little shortcut to getting the eigenvector as 1-1, I might use that shortcut in some of our future examples but the summary of all this is that we have to eigenvalues, 3 and 5.2083

Each one corresponds one eigenvector, the 3 corresponds the 3-1 and the 5 corresponds to 1-1, on our next example we got a 3 x 3 matrix, -1, -2, -2, 1, 2, 1 and -1, -1, 0.2096

We want to find the eigenvalues and eigenvectors of this matrix, this one is going to be quite a bit messier algebraically than the previous example but the principles are still exactly the same.2115

Let me walk you through that, remember we want to figure out A- R(i), now R(i) means you would you have R's down the main diagonal, what we are going to do is we are going to subtract R's from the main diagonal.2126

I'm going to go ahead and write those in, A- R(i) on the original matrix will be -1 - R 2 - R and 0 - R and now we got take the determinant of that matrix, I'm going to highlight things in greens so it does not get too messy here.2140

But it is going to be a little bit complicated because remember for determinant of a 3 x 3 matrix, you got 3-4 diagonals and then 3 back diagonals, it does get a little messy we have plus on each of the four diagonals, minus on each of the back diagonals. 2158

Let us go ahead and figure each one of those out, the first main diagonal is -1 - r x 2 - r x 0 - r which is just -r and now the next diagonal is -2 x 1 x -1 that gives us -2 x 1 x -1.2179

I'm wrapping around the other side, maybe I will go ahead and write those columns to make that a little more obvious, on that first column I got -1 - r, 1 and -1.2207

On the second column I got -2, 2 - r and -1 and if I look that second column, that second diagonal here that is + 2, that gives me a 2 there and I will put a + 2 here.2220

That third column or that third diagonal gives me a +2, I'm going to work back diagonals, that back diagonal right there is negative and I had two other negative, the whole thing is negative.2244

-2× 2 - r, that is that first back diagonal, that second back diagonal again remember that each one of these is negative, or at least we are going to multiply by -1 and then see what it comes out to be.2265

That second back diagonal is -1 - r x -1 and then there is another negative it is plus -1 -r and that third back diagonal is -2x 1 x -r , there are 2 negatives in there and then there is another negative because it is a back diagonal.2281

The whole thing is negative, -2 x r because there is a negative, there is a negative and there is a negative, you put them all together you get three negatives and it simplifies down to a single -2r.2304

We get a fairly complicated expression here, let us expand things out as best we can, this -R I'm just going to write it as -r here and I'm going to multiply these two terms together.2327

I see I got -r, ir is r^2 -2r + r is -r and -1 x 2 is -2, let me collect some terms here, I see I have a +4 that is +2 + 2 - 2 x 2 so -4 + 2r and then -1 -r -2r and again if I multiply this -r through.2339

I see I have -r^3 + r^2 + 2r and let me collect some terms, the +4 -4 cancels, I see I got a -1 and 2r - 2r cancel, but I still have a -r here and if I simplify that all down, I got -r^3 + r^2 + 2r - r is +r -1.2376

Remember you want to set that equal to 0, I'm going to find the determinant of A -R(i) =0, I'm going to set this equal to zero and then I'm going to multiply both sides by -1, I do not really like having a negative sign on the leading term there.2412

I'm going to go r^3 - r^2 - r + 1, I got that by multiplying both sides by -1 there and that is equal to 0, now I'm going to solve this by grouping, I have to somehow factor cubic polynomial and it is usually useful to do that by grouping.2428

I'm going to write this as r^3 - r^2 and then there is this r-1 is equal to zero, in fact that is r^2 x r - 1, -r -1=0 and I get (r^2 - 1) x r -1 factoring out r -1 is equal to zero.2451

But r^2 - 1, I know I can factor that into r -1 x r +1 and when I finally factor this all the way down, I see I have two roots of r=1 and one root of r= -1, those are my eigenvalues for my matrix 1 and -1.2467

Now I still need to go through and find the corresponding eigenvectors of the matrix and I need to find eigenvectors for each one of those, for r=1 and for r= -1.2513

I'm going to start with r=1 actually I think I'm going to start with r= -1 and I'm going to plug that back in to the matrix, I'm going to plug that back into the -r for each one of these.2523

Since r is -1 that means I'm adding one to each one of those entries, let me write that out by adding 1 to each one of those entries -1+ 1 that will give you 0 and I have -2, -2 1 now I have a main diagonal 2 + 1 will give you 3 and 1, -1, 0 + 1 gives me 1.2537

I want to find the eigenvectors corresponding to this matrix which means I have to do row operations on this matrix, I think what I'm going to do is I'm going to bring this third row up to the top and then I'm going to use that to clear out the other rows.2565

I will put 1, 3, 1 on the top row, switch that down to 0, -2, -2 and I'm going to use that third row that 1, 3, 1 row to clear out this -1, I'm going to add one copy of the middle row to the bottom row.2587

These are elementary row operations that we learned how to do in linear algebra, if you are a little rusty on this, I got some lectures on linear algebra here on www.educator.com.2605

You want to check those out and just remind yourself how to do elementary row operation.2614

If I add this 1, 3, 1 to -1, -1, 1 what I'm going to get is 0, 2, 2, that is already looking a little better, I created 0's in my first column, take that one more step, I'm going to get 1, 3, 1 in the first row.2619

In that second row I can divide by -2, I will get 0, 1, 1 and then I can use that 1-1 to clear that bottom row, I could subtract it completely and just get 0, 0.2643

I think I'm going to take it one more step, I'm going to use this 1 to clear away this 3, I'm going to subtract 3 copies of the second row from the first row.2658

That will give me 1, 0 and I subtracted 3 copies so that is 1, 0, -2, 0, 1, 1 and 0, 0, 0, I want to see what x, y, z would have to be in order to give me 0, 0, 0 there.2670

If I look at that, I see that I have several equations here, from the first one I got x + 0y - 2z is = 0, from the second equation I got 0x + y + z=0, I'm going to build my eigenvector and remember how I said in the previous example.2693

You can sometimes start by just taking the last entry to be one, and that works this time, my z is equal to 1 then y + z=0, y would have to be -1 and x - 2z =0, my x would have to be 2.2719

I'm going to fill that in, 2 and then my y is -1, and my z is 1, if you are uncomfortable with that, if you are not quite sure how I did that, what you can do is you can solve this by free parameters.2743

You just set z to be an arbitrary free parameter t and let me go back wards and say since y + z =0, you will get y=-t, since x - 2z=0 you will get x = 2t and you will get x, y, z = 2t/t and you can factor out the t.2755

You will just get 2, -1, 1 x t, if you take t=1, you will end up with the same eigenvector that we found by my little trick of just taking the last entry to be 1.2797

What we got so far is we found the eigenvalues -1 and 1, we found an eigenvector corresponding to one of the eigenvalues for -1, we still need to find an eigenvector corresponding to the other eigenvalue r=1.2813

Let me recap what we did here on this side before we move on and find the eigenvectors corresponding to the other eigenvalue.2830

I will go over this in red here because it looks like I already used up the blue and the green pastel pen. What we first did was we looked at A-R(i), that means we subtracted and r on the main diagonal here.2840

And we found the determinant which meant this cross multiplying 3 main diagonals and 3 back diagonals, the main diagonals each one of them is going to be plus on the main diagonals, the 4 diagonals.2852

When you do the back diagonals, each one is going to be negative, you are multiplying positively all these 4 diagonals, that is what we are doing here and then multiplying with an extra negative factor all these back diagonals.2868

That gets to be messy and you have to be very careful not to make a mistake as you work through all those but it simplifies down, we worked out all the algebra here and simplify it all the way down to r^3 + r^2 + r -1.2886

We multiply everything here by -1, both sides by -1 in order to get a nice, positive coefficient on the r^3, then we could group the terms and factor out r - 1 on each term.2903

We factor out r - 1 by the whole thing, r^2 - 1 that factored down into r + 1 r- 1 and when we looked at the factors we got a double root at r=1 and a single root at r=-1, those are my 2 eigenvalues 1 and -1.2917

I'm going to find eigenvectors for each one, r=-1 we plugged that back into the matrix which means we are subtracting -1 along the main diagonal which effectively means we are adding 1 on the main diagonals.2939

That is where we got the 0, 3, 1 on the main diagonal, all the other entries are just the original matrix, we are row reducing this according to our rules for linear algebra and we row reduced it to reduced row echelon form.2954

From there we are trying to solve for x, y, and z, you can use free parameters the same way you learned in linear algebra or you can use my little trick of starting with z=1 and then working backwards to figure out what x and y should be.2968

Either way you end up with this as the eigenvector and remember that is just for the eigenvalue r=1, we still need to go back and find the eigenvector corresponding to the eigenvalue r=1, that is what we are going to do on the next slide.2985

That should have been an r, our eigenvalues are r 1 and -1, we already found the eigenvectors corresponding to -1, that one is already done on the previous side, you can go back and check that out if you are just jumping to this part.3012

We still need to find the eigenvectors corresponding to r=1, the way we do that is we plug in A-R(i), which means you subtract r on the main diagonal, I will do -1 there, -1 there, and -1 there.3032

I will see how that works out, A-R(i) is equal to, now -1 minus -1 is -2, -2, -2, 1, 2-1 is 1, 1 there, -1, -1, 0-1 is -1.3049

What I see is basically 3 copies of the same row, they are all multiples of 1, 1, 1, if I pull that 1, 1, 1 up to the top I can get 1,1, 1 in the top row and then I can use that and multiples of that row to kill all the other rows, send all the other rows to 0.3069

I'm trying to find vectors x, y, z, that when I multiply that x, y, z I get 0, all that really tells me is that x + y + z=0, I really do not have very much information about x, y, and z.3091

I'm going to use my free parameters to solve that, I'm going to set up z=t, I still do not have enough information to solve for y, I'm going to set up another free parameter s there.3108

Now I can solve for x because I have x + y + z=0, that is telling me that x + s + t=0, my x would be -s -t.3123

My x, y, z vector is equal to, my x was -s -t, my y was s and my z was t, what I like to do is factor that out and get an s part and a t part.3140

I'm going to get an s vector and I'm going to get a t vector, I see the coefficients of s are -1, 1, and 0, the coefficients of t here are -1, and there is no t in the middle so 0, there is a 1t in the bottom.3160

The vectors that I get are s x -1, 1, 0, and t x -1, 0, 1, my eigenvectors, looks like I have 2 independent eigenvectors, if I just take different combination of s and t.3186

These are corresponding to the eigenvalue r=1, my eigenvectors are -1, 1, and 0, and -1, 0, and 1.3209

I'm finally finished with finding eigenvalues and eigenvectors there, but let me recap how we worked all that out.3244

The first thing we did in this problem was on the previous side, but we started out with A-R(i) and we took its determinant, that was a bit messy and we ended up subtracting r's on the main diagonal.3255

And expanding out a 3 x 3 determinant which meant multiplying 3 things down the positive diagonal directions and then 3 negative diagonal directions.3267

It is quite messy and we got a cubic polynomial and then we solve that out, we got r=1 or -1 and on the previous side we plugged r=1 back in and we found an eigenvector on the previous side.3282

That one worked out ok, what we are doing here is we are finding our eigenvectors for r=1, we are plugging in r=1 into A-R(i) here, we subtract 1 down the main diagonal, that gives us this matrix.3309

When you look at this matrix, we really have 3 copies of the same row because they are all multiples of 1, 1, 1, that means we can use the 1, 1, 1 row to wipe out the other 2 rows.3327

We just end up with the equation x + y + z=0, that does not give us very much information about x, y, and z, we have to use free parameters to solve it.3339

To start out, we always work backwards from z, I will start out with z = t, then I still cannot figure out what y is, I set y = s and then finally if I know what s and t are, I can solve for x in terms of t and s.3350

x= -s -t, if I plug those back in, the x was -s -t, y was s, the z was , that is what I figured out over here, plugging those back in.3363

Then I can factor out in to an s vector and t vector, I factor out the coefficients of s here and the coefficients of t here.3377

Those are 2 independent eigenvectors corresponding to r=1, those are the 2 eigenvectors that I associate with the eigenvalue r=1.3386

I found all the eigenvalues and eigenvectors of this matrix, that is the end of our review of linear algebra lecture here on the differential equation series on www.educator.com.3397

If that was not enough linear algebra to get you completely comfortable with finding eigenvalues and eigenvectors, what you might want to do is go and look at the linear algebra lectures here on www.educator.com.3409

You will find a lot more examples and a lot of this stuff work out in a lot more detail, what we are going to do in the differential equation lectures is we are going to start using some of this ideas to solve systems of differential equations.3417

On the next lecture we will look at systems of differential equations, we will write them in matrix form and then we will use eigenvalues and eigenvectors to solve them.3431

That is what we have to look forward to next on www.educator.com.3441

My name is Will Murray and you have been watching the differential equations lectures. Thanks a lot for joining us, bye.3444

Hi there and welcome back to www.educator.com and these are the differential equations lectures, my name is Will Murray.0000

Today we are going to talk about systems of equations and the way you solve systems of equations is essentially by finding the eigenvalues of a matrix, there is a sort of three different things that can happen.0007

It depends on what kind of eigenvalues you get when you solve for those eigenvalues in the matrix, sometimes you get real eigenvalues but they are repeated and sometimes you get complex eigenvalues.0019

What we are going to do is we will have a separate lecture for distinct real eigenvalues when you get different real eigenvalues and then we will have another lecture for complex eigenvalues and then we will have another one for repeated real eigenvalues.0033

Each one of those kind of has its own separate behavior, what we are talking about today is distinct real eigenvalues, if you are working on some homework problems and you are finding that you are getting complex eigenvalues or repeated eigenvalues.0046

Then we got separate lectures to cover those cases, you might want to check out those lectures instead, today we are talking about distinct real eigenvalues, let us jump in and see how that works out.0061

We are going to solve the system of linear differential equations x1(t) is equal to a1-1 x1 + a1,2 x2 and x2′(t)=2,1 x1, a2, 2 x2(t) that is the system we are trying to solve.0072

Right away we are going to try to write that in matrix form, first of all I'm going to drop off the t's on the left hand side and I'm going to try to simplify this equation, I simplify it into that form that really has not changed much.0095

Then I'm going to try write it in matrix form which means I'm just going to extract the coefficients (a1,1), (a1,2) (a2,1), (a2,2) and write those in matrix and then I'm going to write x1 x2 as a column vector and the derivatives of those as a column vector on the left.0110

The shorthand version of this whole thing is to write this 2 x 2 matrix as a and then this x stands for the column vector x1 x2 and the x′ stands for its derivatives, that is bold vector x standing for (x1 x2).0130

Bold vector x′ standing for its derivative, this X′ equals (ax) in bold form that very small equation there is just short hand for this entire system and we are going to figure out how to solve that.0151

Let us go ahead and see how that works out, the secret to solving that is to look at the matrix a and find its eigenvalues and their corresponding eigenvectors.0168

It all comes down to the eigenvalues and eigenvectors, now if you are a little rusty on your linear algebra, we had of lecture here on the differential equation series which was a review of linear algebra.0179

We practice finding eigenvalues and eigenvectors, if you are little rusty on that, if you do not remember that, why don't you go back and look at the review of linear algebra around the lecture, it was the previous one here in the differential equation series.0190

You will get some practice calculating eigenvalues and eigenvectors then come back to this lecture, we are going to use it to solve differential equations, we are going to find the eigenvalues, we are going to call those r1 and r2.0204

Each one of those is going to come along with the corresponding eigenvector and we are learning a new Greek letter here, this is called xi 1, it is hard to say, it sounds like KSI, KSI is the way you pronounce that.0217

We write it as a backwards three, it is a complicated but with that is the Greek letter that we are going to use for the eigenvector, fortunately you would not have to write it very often.0235

But we will call the eigenvector xi 1 and xi 2, xi 1 is the eigenvector corresponding to the eigenvalue r1 and xi 2 is the eigenvector corresponding to the eigenvalue r2.0249

Once you find those eigenvalues and eigenvectors you can write down the general solution to the differential equation right away, the general solution is just a constant times that first eigenvectors c1 times e^r1(t)remember r1 was the eigenvalue.0265

And then a constant times the second eigenvector times e^r2(t), basically it is just a matter of finding the eigenvalues and eigenvectors and then just dropping them into the solution formula.0283

The initial conditions there is a small mistake here on my print out here, the initial conditions are something we would use to find the c1 and c2, we use the initial conditions that were given to find c1 and c2 here.0298

Let us see what else we want to do with those, the next step you take with that would be to graph the solutions and let me give you a quick idea of that and I'm probably make more sense after we study some examples.0318

What we will have is a set of axis, this is X1 in this direction and x2 in this direction, and we will have 2 vectors c1 and c2, first I want to graph those vectors c1 and c2.0339

Maybe you have a vector there which is c1, it got a little messy there and then maybe in another direction you will have c2, and what you want to do is to set up axis defined by those lines .0355

I'm going to set my axis up, define by those lines by the eigenvectors and the solution trajectories, now remember that they all had this term of e^r1(t) and e^r2(t).0385

The solution trajectories will either go to zero or infinity depending on whether the r1 and r2 are positive or negative.0407

If r1 is positive then we will have trajectories on the c1 axis going out to infinity, if r1 is negative then we would reverse these arrows and have them going in toward 0.0415

For r2 again if it positive and xi2 axis we would have solution trajectories going out towards infinity, if it is negative I will graph it as if they are negative then these solution trajectories would tend in toward zero.0422

We also have solution trajectories in between these two axis and what happens with the ones in between is they tend toward the axis spanned by the eigenvector that corresponds to the larger eigenvalue.0447

You look at r1 and r2 and you see which one is bigger and then you have the solution trajectories tending towards that axis and the way I graphed it right here, I graphed it as if r1 is positive because those are going off to infinity.0461

I graphed it as if r2 was negative because I graphed them going into zero, what that would mean is that the solution trajectories in between these axis would tend towards the axis defined by the r1 because that is the larger one.0477

They follow these these curves that would follow the other axis for little way but then get drawn off in the direction of the r1 axis, you see how all these red curves are tending towards this axis determined by the first eigenvector there.0495

All these axis, all these solution trajectories are tending towards this southwest to northeast axis, they do not cross each other by the way, we never have solution trajectories crossing each other.0520

We will get some practice of that, I think it is probably hard to see in the abstract and when practice several examples you will start get the hang of it, let us go ahead and start our first example here.0535

We have to solve the following system, this is actually an initial value problem solution, there are going to be two parts of this, first is solving the differential equation which is of the form X′ = ax.0548

We will have some initial conditions which we are not going to worry about until later, let us set up the the differential equation first, now remember that the solution to all of these things depends on finding the eigenvalues and eigenvectors of the matrix.0561

Let us find the eigenvalues and eigenvectors of the matrix, A-R(i) , this is how you find the eigenvalues is 6 - R - 2, 9 - R and we want to take the determinant of that.0582

The determinant we cross multiply is 6 - R x 9 - R - 2 x -2, that is - + 4, -4, we will get R^2 here - 9R - 6R, -15R + 54 - 4 is + 50=0.0600

That factors really nicely, that factors into R -5 times R -10 is equal to 0, we get our eigenvalues our R is 5 and 10, two eigenvalues for each one of those, we are going to throw those back in to the matrix and find an eigenvector.0632

For our R=5 I will do A- R(i) A-5(i), I will subtract 5 from the main diagonal, 6-5 is 1, -2 -2, 9-5 is 4 and I can simplify that matrix using my row operations, going through the linear algebra kind of quickly.0654

If you are little rusty on this linear algebra we do have a whole set of lectures on linear algebra here on www.educator.com and I also gave one quick review of linear algebra lecture just before this one.0677

You might want to go back and check on that if you want to get a quick refresher for linear algebra if you are really rusty and you want to do it from scratch.0689

We got a whole set of different of linear algebra lectures that you can check those out and get a lot of practice on finding eigenvalues and eigenvectors.0697

What I'm going to do here is do some row reduction, (1, -2) and then I can use that row to cancel out the second row of 0's here and I want to find a vector that will give me 0 when I multiply that matrix by it.0706

My standard trick and I talked about this last time in the review of linear algebra lecture is to put a 1 in for the z component or for the Y component and then figure out what I should have for the X component.0724

It looks to me like I have -2 is equal to 0, that first entry that will be a 2, if you are not comfortable with that method you can use the free parameter method, maybe I will just show that quickly.0738

We got x -2y=0, we use a free parameter for y, we use y=t and then x would have to be 2t, xy would be 2t and t, if we factor out the t we would have 2 and 1 x t.0750

That is another way to get that factor but I usually use my short hand trick of just throwing a 1 in a y position and then figuring out what the exposition would have to be so that they multiply to be 0.0771

That is my little shortcut of finding an eigenvectors, we still have to find the eigenvector corresponding to R=10, what we just did was fighting eigenvector corresponding to R=5.0783

Let us do the same thing for R=10, we subtract 10 down the main diagonal A-10(i) is 6-10 is =4, -2 -2, 9 - 10 is -1 and if we will reduce that then that row reduces into, I could bring the bottom row up to top and maybe multiply it by -1 to get rid of negatives.0795

(2, 1) and I can use that to simplify the first row as well, I will simplify that into 0 and if I do my normal trick of putting one in a y position, 2 x 1 you would just give me 1/2 in the x position and I do not really like that.0828

Remember you can scale eigenvectors if you want and they will still be eigenvectors, I'm going to scale that eigenvector by a factor of 2 and get (1, 2) and I see I made a small mistake there, that 1/2 should have been negative.0847

The whole point is that when we multiply this matrix by this vector, this should be 0, that 1/2 should have been negative in order to make that come out to be 0.0868

When we scale it up by, I will multiply that by 2 just get rid of the fraction, I will put a 2 in the y component and a -1 in the x component, that is my eigenvector corresponding to the eigenvalue R=10.0879

Remember once you have the eigenvalues and eigenvectors there is a generic solution that always works for the general solution to these differential equations, I will remind you what that was.0900

We had that in the beginning of the lecture but it is c1 x e^r1(t), first eigenvector x e^eigenvalue(t) + c2 x c2 x e^ r2(t), what I'm going to do is just drop those eigenvalues and eigenvectors that I figured out into this generic solution.0910

My generic solution here is c1, now my first eigenvector was (2, 1) x e^r1(t) that is e^5t + c2 x my second eigenvector is (-1, 2) x e^10t that is my general solution.0941

If I did not have initial conditions this is where I would stop and say that is the complete solution to the problem, but now I'm going to use the initial conditions which is the x(0)=(0, 5).0970

I have not use that for anything yet, I'm going to incorporate that, that means I plug in 0 for t, I get c1 x (2, 1), e^5 x 0, I'm plugging in e^0 here.0982

I also plug it in over there later, e^5 x 0 is just e^0, that is just 1, c1 x (2, 1) + c2 x (-1, 2), again e^10 x 0 is just 1, that is supposed to be equal to 0× 5.1005

I'm going to separate that out and get two equations and 2 unknowns for C1 and C2, it looks like from the first row I have 2c1 - c2 is equal to 0 and from the second row I have c1 +2c2 is equal to 5.1026

Now you can solve this using whatever technique you like best from algebra, let me use substitution, it looks like c2 is equal to 2c1, I will plug that in and I get c1 + 2 x 2c1 is equal to 5.1048

It looks like I get 5c1 =5, c1 = 1 and from c2 =2c1, I get c2 is equal to 2, I will take those two constants and plug those back into my general solution, I will get my particular solution x=1, that is my c1 (2, 1) e^5 + 2 x -(1, 2) x e^10t.1067

If you wanted you can bring that to into the vector, multiply it by each of the components, does not really make much difference so I'm not going to do that.1107

Now I have finished solving the initial value problem but let me go back and show you what we did with each step there.1120

The whole key to solving this systems of differential equations is to start with the matrix and find the eigenvalues and eigenvectors, that is what we are doing here, we found A- R(i), that means you subtract R along the main diagonal.1127

Take the determinant, that means you cross multiply and we set that equal to 0, we get polynomial which is this one turns to be easy to factor and we got 2 values for R.1140

For each one, you plug those back in, you subtract them back along the main diagonal that one came from a 6 - 5 and that 4 right there came from that 9 - 5, that is using the eigenvalue R=5.1157

We will row reduce that matrix using the techniques we learned in linear algebra for row reduction and I just guess the vector, I started out with the 1 in the y component and then I figured out that the x component would have to be a 2.1169

If you do not like that technique you could also use free parameters where you start out with y=t and you go back to x=2t and then you figure out again that your eigenvectors (2, 1) do the same kind of thing for R=10, the other eigenvalue.1186

Plug it in on the main diagonal, reduce the matrix and then try to find an eigenvector where when you multiply the matrix by that eigenvector you will get zero, our eigenvector there is (-1 and 2).1202

We just drop those eigenvalues and eigenvectors into our general solution, those are the eigenvectors and those are the eigenvalues right there and that is our general solution, we will be done except the we also have to satisfy this initial condition.1216

The initial condition we use it right here and that means we are plugging in t=0 which makes the E terms dropout, you just get e^0 is 1 and we end up getting 2 equations for 2 unknowns in the c's which is pretty easy to solve.1234

It is just kind of high school algebra there to solve the 2 equations and 2 unknowns, get values for c, plug those back in there is c1 and there is c2, we got 1 and 2 and we have our particular solution to the differential equation.1249

In the next example we are to going to practice graphing some solution trajectories to the general system to the general solution for the previous system, let us see how that works out.1267

Remember, there is a sort of several stages here, you start out by graphing the eigenvectors, I'm going to make a big graph here. 1280

First I will graph the eigenvectors and we will see what the axis are defined by those eigenvectors, let me put a scale on here.1292

That first eigenvector is (2,1) and remember you think of that as being X and Y are X1 and x2, your x1 is 2, x2 is 1 or x is 2 and y is 1, I'm going to go over to (2, 1).1313

There it is right there, there is the vector (2, 1), 2 horizontally and 1 vertically and I will graph negative multiples of that and I'm going to set up an axis defined by that eigenvector, I will do that in blue.1331

I see that I have a positive eigenvalue there, the eigenvalues 5 that is positive which means solutions on that axis are going to go off to infinity in the positive direction and negative infinity in the negative direction.1353

That really depends on whether C1 is positive or negative, this is kind of C1 is positive, this C1 is negative that tells me whether solutions go to positive infinity or negative infinity.1370

I'm going to set up the other eigenvector as (-1, 2) so -1 on the horizontal axis, 2 on the vertical axis, there is (-1, 2 ) and let me set up my axis, I wanted to do that in blue.1382

I see also that I have a positive eigenvalue there, R+10 my solutions are also going to infinity in the positive direction along that axis, that is c2 greater than zero because that is (-1, 2), this is c2 less than 0.1415

All of my solutions seem to be going to infinity that is because I have 2 positive eigenvalues but now let me look at solutions in between there, I know that all the solutions go to infinity but if I mix these two solutions what I notice is that the e^10t is the dominant term.1437

That is because that is the larger eigenvalues this is really key here, 10 is bigger than 5 which means that mix solutions in between these axis are going to tend towards the c2 axis.1457

The axis defined by the larger eigenvalues, if I start in between these solutions my other solution trajectories are going to bend over and I'm going to pull towards the axis defined by the larger eigenvalue.1475

That is why all of these solutions in between are going to approach the direction defined by the larger eigenvalue that e^10t, they all bend towards the axis defined by the vector with the e^10t.1495

Let us recap what we did there, we got a nice graph of solutions but how do we get there, the first thing was to graph the two eigenvectors and set up axis there, (2, 1) that was my vector right here, there is (2, 1).1523

Here is (-1, 2) and each one of those gives us an axis of solutions and both of those have positive eigenvalues which means we are going to infinity along both of those axis, that is why these arrows here go outwards.1540

And why these arrows go outwards and then when we look at solutions in between, they want to pull towards the axis determined by the larger eigenvalue, let me just remind you this was the eigenvalue R=10 this is the eigenvalue R=5.1559

All the mix solutions are going to pull towards the R=10 axis and that is why all these curves tend towards the c2 axis, the axis defined by that eigenvalue of R=10.1578

In example 3 we are going to solve another system of linear differential equations, again remember the whole key to solving these things is to find the eigenvalues and the eigenvectors of the matrix.1599

Let us jump right in and find the eigenvalues of this matrix, we want to find A- R(i), we are going to subtract R's down the main diagonal, 3 - R2 (-3 and -4 - R), I will multiply that out, 3 - R x -4 - R, - -3 x 2, that is + 6 = 0.1615

If I work that out I will get r^2 + 4R -3r, r^2 + r and then -12 + 6, that is -6 is equal to 0 and that is a nice format factors very nicely into R +3 and R -2=0.1646

I get my eigenvalues very nice, R=2 and -3, that is the first step is getting the eigenvalues but then remember for each one of those you are going to find an eigenvector, let me start out with R=2 and find my eigenvectors there.1672

That means if I plug in A- R(i), I get 3 - 2 is 1, 2, -3 - 4 -2 is -6 and I can quickly row reduce that, the 1, 2 will cancel out the -3 -6 I get 1, 2,0, 0 and I'm going to use my trick of putting a 1 and Y component.1688

I see that I would have to put a -2 in the X component in order to get zero multiplying those out, if you do not like that trick, you can use free parameters to get the eigenvector but I can see there that my eigenvector is (-2, 1).1716

For R=-3, that was my other eigenvalue, again I'm going to find A- R(i) that means I'm going to add 3 along the main diagonal, 3 + 3 is 6, 2 - 3 now -4 + 3 is -1 .1736

I see that I can simplify that a bit, I can divide the top row by 3 and then add it to the bottom row to cancel out the bottom row and I see that if I put a 1 in the y component, then my X component, in order to make it cancel would have to be -1/3.1756

I do not really like having a fraction in there, it is a little easier to have whole numbers, I multiplied both entries there by 3, get rid of that 3 in the denominator and you can do that with eigenvectors.1777

Remember any scaler multiple of an eigenvector is still an eigenvector, then I will get -3, 1, I'm sorry (-1, 3) that is my eigenvector for corresponding to R= -3 and now I can find my general solution.1795

Remember my general solution that is always the same, we saw the general form of the solution at the beginning of the lecture in the overview, maybe go back check that out if you are not sure where this is coming from.1817

But it C1 times the first eigenvector xi1 e^r1(t), r1 is the first eigenvalue plus a constant times a second eigenvector xi2 times e^r2(t), R2 being the second eigenvalue, my general solution in this case is C1 first eigenvector is (-2, 1).1831

e^first eigenvalue is 2t + c2, second eigenvector is (-1, 3) and e^-3t, that is my general solution but I also got this initial condition that I want to satisfy, I want to plug that in, I will plug in t=0 then I'm going get e^0 twice here.1860

Both of those terms turn into 1 and they dropout, I will get C1 x (-2, 1) + c2 x (-1, 3) = it is supposed to be (5, 0), let me set that up.1888

That will give me a system of 2 equations and 2 unknown, it should not be too bad, -2 c1 - c2= 5 and c1 + 3c2 = 0, we can solve this out using whatever algebraic system you are comfortable with.1910

But I'm going to use a substitution, I see I'm going to use the C1 as -3 C2 and if I plug that into the first equation I'm going to get 6C2 because I'm plugging them in right here, -3 x -2 is 6, 6c2 - c2 is 5.1932

5c2=5, c2=1 if I plug that back and I get C1 i= -3, I can drop those two constants into my general solution, now my general solution is c1 is -3 x where was my eigenvector (-2, 1) times e^2t + C2 that is just 1 x my eigenvector is (-1, 3) x e^-3t.1957

That is not a general solution anymore because we already found the constants, that is my solution to the initial value problem.2007

That is the end of that problem but let us recap and see the steps that went into solving that, the first step to solving systems of differential equations is to find the eigenvalues and the eigenvectors of the corresponding matrix.2025

That is what we are doing here, we are finding A- R(i), we are subtracting R's down the main diagonal, take the determinant that means cross multiply pluses and minuses here and we get a fairly simple quadratic equation for R which solve quickly into (2, -3).2039

That means those are the two eigenvalues, but for each one of those we have to find an eigenvector, we plug the first one back in R=2, we are subtracting -R down the main diagonal, that is where we get a 1 here and a -6 here.2057

By row operations that simplifies down to 1, 2, 0, 0 and we are looking for vector that matrix would kill and if we just start with a 1 and a y component, then it is pretty easy to see the X has to be a -2.2076

You can also use a free parameters to solve that if you are more comfortable with that technique, then we try to do the same thing with R= -3, that means we are adding 3 to the main diagonal of the original matrix.2090

And (3, 1), if we start out with a 1 and Y component we get -1/3 in the X component, which I do not like very much because of the fraction, you can scale up eigenvectors if you like,I'm going to scale that up to (-1, 3).2105

We drop those eigenvalues and eigenvectors into the general solution, the eigenvalues are (2, -3), the eigenvectors are (-2, 1) and (-1, 3) and that would be the end of finding the general solution but we also got this initial condition here.2119

Using the initial condition means we are plug in T=0, the E terms you get e^0, those turn into1 and then we have C1 times a vector plus C2 times a vector is equal to this given vector (5, 0).2138

That just resolves into a system of two equations and two unknowns which you can solve however using whatever technique your most comfortable with, you can use substitution to get a value of C1 and C2.2154

And I drop them back into my equations my C1 was -3 and my C2 just came in there as a 1 and I had my complete solution there, hang on to this general solution.2170

We are going to come back and look at that general solution in the next example and see how that would graph.2188

Let us go ahead and move on to the next example and try to draw some solution trajectories for that general solution. In example 4 we are going to graph some solution trajectories to the previous system of equations.2194

We already found the general solution back in example 3, this came from example 3, if you did not just watch example 3, you might want go back and look at example 3 and you will see where this solution came from.2208

What we are going to do in this problem is we are going to see what kind of solution trajectories that would give us, I'm going to set up some axis here and we are going to draw some solution trajectories and try to get a feel for the behavior of some of those solution.2223

This represents X1 and X2 or you can think of them as X and Y, it does not really matter and I'm putting a scaler here.2242

Now remember the first step here is to graph the eigenvectors and to set up 2 axis corresponding to the eigenvectors, our first eigenvector is (-2, 1), I'm going to go over to -2 and then up one unit and there is my eigenvector (-2, 1).2262

I'm going to set up an axis spanned by that eigenvector, let me see where to be on the negative side, right there, there is my axis defined by the first eigenvector.2282

My second eigenvector is (-1, 3), there is -1 on the horizontal axis, 3 on the vertical axis, it is like it is up there and its counterpart in the negative direction will be down there, I'm going to set up an axis corresponding to that.2302

Those are my two axis corresponding to the two eigenvectors, you can think of this as being the the C1 axis corresponding to the first solution c1 and there is the C2 axis corresponding to the second solution there.2328

I will fill in my axis a little bit and now let us try and figure out whether those solutions are going to zero or infinity and that really comes down to the sign of the eigenvalue, if the eigenvalue is positive then it goes to infinity.2346

If it is negative it goes to zero, let us look right here, we got a positive eigenvalue for C1, that means the solutions on that axis are going to infinity, I'm going to draw these axis going out to infinity.2362

On the second axis, we got a negative eigenvalue that means those solutions are dying down to zero, that tells us which direction everything is going along those two axis.2378

Finally our mix solutions will be in between these two axes and I always approach the axis defined by the larger eigenvalue. 2399

Here are two eigenvalues are 2 and -3, what we see is the 2 is the larger eigenvalue right there, that means the mix solutions will approach the C1 axis, which is this axis right here.2410

The mix solutions are all going to tend asymptotically towards the C1 axis, they might follow the C2 axis for a while but they all end up approaching the C1 axis.2431

That is because we got a larger eigenvalue for that solution, you see how I'm kind of drawing the arrows, drawing the lines to make them parallel to the arrows on the two axis defined by the eigenvalue.2451

But they are always heading towards the one defined by the larger eigenvalue which is the C1 axis because it had an eigenvalue 2, the other one was negative.2470

Let me recap how we solve that one, we got a nice graph of solutions here but the way we did it was by starting out by graphing the eigenvectors, I graphed that first eigenvector it is right there is (-2, 1).2485

-2 horizontally and 1 vertically, then I graph that second eigenvector was (-1, 3), there it is right there, it is that black dot right here is (-1, 3) and then I set up axis based on those two eigenvectors.2500

I drew my blue lines here, those are my axis based on those two eigenvectors and then I looked at whether the eigenvalues were positive or negative to see whether solutions are expanding or contracting along those axis.2516

What I see is that the first axis is positive, it has a positive eigenvalue, that means solutions are expanding along this axis, C1 I drew my arrows outwards, all those arrows are going outwards along the C1 axis because the solutions are expanding.2529

On C2 I got a negative eigenvalue which means the solutions are contracting along that axis, that is why you see these arrows are going in towards the origin we are on that axis.2546

I want to draw my mix solutions, what I do is I looked at which of the eigenvalues is bigger, well 2 is definitely bigger the -3, that means the C1 axis is going to dominate, which means all these mix solutions tend towards the C1 axis there.2559

They are parallel to the C2 axis may be at the beginning but then they end up tending towards the C1 axis, because C1 had the larger eigenvalue.2578

That is how you can fill in solution trajectories to cover the whole plane, remember they never cross each other which you can see what the behavior of any solution trajectories starting at any point in the plane would be.2591

That is the end of that example, let us move on to example 5 where we are going to solve a system that looks a bit like one we solved earlier today but it got some important sign differences, let us see how that works out.2604

We are going to solve the system (-6, 2), (2, -9) as our matrix, remember on all of these the first thing we do is find the eigenvalues and the eigenvectors of the matrix, let us jump right in there.2616

A- R(i) means I will subtract R down the main diagonal, -6 - R2, 2 -9 - R and I want to take the determinant of that, I will cross multiply -6 -R x -9 -R, -2 x 2 is -4 and I want to that equal to 0, if I expand out the algebra I see I get a negative on both places.2630

That will give me a positive R^2 -6 x -R that is + 6R, same with a 9, so + 9R + 15R and then -6 and -9 multiply to positive 54 but -4 gives me a + 50 is equal to 0 and that factors really nicely into R +5 and R +10.2660

Of course that is easy to solve, my two eigenvalues are R= -5 and R= -10 and if that means for each one of those I'm going to plug it back in and find the corresponding eigenvectors, for R= -5 I'm going to start with that one, R=-5.2694

My A- R(i) is the same as adding 5 to each entry on the main diagonal, -6 + 5 is -1, I will put a 2 there, -9 + 5 is -4 and if I do a little matrix reduction, I'm going to take that first row and multiply it by -1, I get (1, -2) in the first row.2714

I can use that to completely wipe out the second row, I will just get 0 there and to find the eigenvector I have this little trick of writing a and y component and then seeing what the x component would have to be.2744

If they multiply to be zero, I see in this case the x component would have to be 2 to multiply the 0, if you are not comfortable with that trick just use free parameters to solve this, the same way you learned how to do a linear algebra class.2759

If you want some review on that I have a quick review on linear algebra in the differential equation series, it is a lecture right before this one and you can get a longer review on linear algebra with the whole separate linear algebra series here on www.educator.com.2772

My eigenvector(2, 1) right there and now let us find one for R= -10, that means I'm going to add 10 to each element on the main diagonal to get A- R(i) is -6 + 10 is (4, 2), 2, -9 + 10 is 1, I can row reduce that matrix.2787

I'm going to chop the first row down by a factor of 2, I get (2, 1) and then I will use that to wipe out the second row, I get (0, 0) and again I'm going to put a 1 in the y component.2819

That looks like I will have to put a -1/2 in the x component to make that come out to be 0, I'm not that comfortable with a fraction but remember you can multiply eigenvectors by a scaler and you will still get eigenvectors.2831

I'm going to pick 2 for my scaler and I will get that -1 and 2 as my eigenvector corresponding to the eigenvalue R= -10 and remember my general solution gave this at the beginning of the lecture in the lesson overview it is always the same for the systems.2849

It is always C1 times your first eigenvector xi1 times e^ri(t), r1 is the first eigenvalue + C2 x C2 x e^r2(t) and I drop in the xi1 and xi2, those are my eigenvalues and eigenvectors.2855

My general solution is C1, my first eigenvector was (2, 1) and my first eigenvalue was -5t and my second eigenvector was (-1, 2) x e^-10t and ordinarily at this point if I had some initial conditions I will be using them right now to figure out what my C1 and C2 are.2900

But I was not given any initial conditions in this problem, I'm just going to stop there and say that my general solution is the best that I can do, I will present that as my solution there.2931

That is the end of that one, what we are going to do in the next example is come back and graph some solution trajectories based on this general solution but before we go ahead and do that, let us recap what we did to get to this general solution.2946

We started with the matrix and we got find the eigenvalues and eigenvectors, to find the eigenvalues you subtract R off the main diagonal and then you take the determinant by cross multiplying positive in that direction and negative in the back direction.2959

That gives you this quadratic polynomial which factor really easily, R was -5 and 10, if you plug those values back in R= -5 means you are adding 5 to the main diagonal, that -6 + 5 that is how I got this -1 right here and -9 + 5 + 5 is how I got this -4.2975

I row reduced that, which meant I could wipe out the bottom row, simplify the first row and then I was looking for an eigenvector which means I am looking for X and Y that are killed by that matrix.3001

By starting out with a y=1 I was able to guess fairly easily that x had to be 2. If you do not like that kind of guessing technique, use free parameters. It is a very systematic technique and it will get you to the same answer.3014

That was my eigenvector corresponding to R= -5 I did the same thing with R= -10, I added 10 to the main diagonal, simplify down the matrix and I got an eigenvector, I did not like that one because it had a fraction it so I multiply that by 2.3030

You can do that with an eigenvector, you can scale it by something if it makes it more convenient, I was trying to get rid of that fraction and I got this new eigenvector, I took my two answers and I drop my eigenvectors into my general solution.3048

I drop my eigenvalues the -5 -10 into the general solution and I got my general solution and for this problem that is as far as we can take it because we do not have an initial condition.3062

There is no way to solve for the C1 and C2, that is the end of this problem but hang on to this general solution, we are going to use it in the next example to draw some solution trajectories for this.3073

You will be wanting to remember this general solution, we are going to use it right away in the next example, in example 6 we are going to graph some solution trajectories to the previous system of equations.3087

This is the general solution that we figured out in the previous system, if you are not sure where this comes from, go back and rewatch example 5 if that was where we figured out this general solution, this is coming straight from example 5.3099

We started out with a matrix there and we found its eigenvalues and eigenvectors and that is how we got this general solution, what we are doing in this example is we are going to try to graph this.3115

I'm going to set up some axis here and you want to start with your eigenvalues and eigenvectors, here is x1 and x2, I'm going to give myself some scale.3127

You want to start out by graphing your eigenvalues and eigenvectors, first eigenvector I see is (2, 1), I'm going to graph that right here horizontal 2, vertical 1 on the other direction and I'm to go ahead and set up an axis there.3156

Corresponding to the first solution there, that is my C1 axis corresponding to that first solution, I'm going to look at the second eigenvector which is (-1, 2) .3177

-1 in the horizontal direction, 2 in the vertical direction, there it is right there, there it is negative, let me set up an axis coming down through there.3192

I want to figure out which direction the solutions are traveling along those axis and the way I do that is by checking the eigenvalues and seeing whether positive or negative.3214

I see on my C1 solution I got a negative eigenvalues that means that the solutions are shrinking there, they are getting smaller and smaller, all along this axis I'm going to show my solutions gradually drifting into 0.3229

My second one, I see the eigenvalue there is negative 10, that means the solutions are also getting smaller on the second axis, all of these solutions are drifting into 0 and I want to think about my mixed solutions where I start mixing the C1 solutions and the C2 solutions.3248

The mixed solutions are always dominated by the larger eigenvalue, the mixed solutions I'm going to look for the larger eigenvalue, and my 2 eigenvalues are -5 and -10, you want to be really careful here.3270

-5 is larger than -10 because it is bigger on the number line than -10, it is -5, that means that is the dominant solution here, the dominant axis is the one defined by C1, that means all my solutions in between these 2 blue axis will try to tend towards the axis defined by C1.3284

I'm going to draw these solutions in between here, they are all going to 0 because everything here goes to 0, but they are going to 0 along the axis defined by C1.3323

All of these things are going to try to follow the axis defined by C1, even the ones that start near C2 are ultimately head over and try to follow the solution defined by C1.3336

The reason for that is because C1 has the larger eigenvalue even though -5 and -10 they are both negative, they are both very small but you still want to find the larger one and -5 is larger than -10, let me emphasize that.3365

We are using the fact that -5 is larger the -10, -5 gets to call the shots on the other solutions, all the other solutions will be attracted towards that larger eigenvalue, it is essentially means that e^-10t dies out quickly and the e^-5t is the dominant term there.3382

That is our graph of solutions. Let me recap the steps to doing that. First of all we took the general solution and we took that back from example 5.3406

If you are wondering where this line comes from, maybe go back and watch example 5 and you will see we did all the arithmetic to work out those eigenvalues and eigenvectors.3415

This problem we are just graphing them, we start out with the first eigenvector (2, 1), we graph (2, 1) right there that is 2 on the horizontal, 1 on the vertical that is where we got that first black dot there.3423

We looked at (-1, 2) and we graph that -1 horizontal, 2 vertical, graph that point as that second black dot and then we use those vectors to define our 2 axis, that is where we got these two blue lines.3439

We had to figure out are the solutions along those lines expanding or contracting and we figure that out is we look at the eigenvalues, here we see a negative eigenvalue which means our solutions are contracting.3457

Which is why we had these arrows going into zero all along that line, here we have also a negative eigenvalue again we have solutions contracting, they are going into 0 all along that line.3473

Our solutions in between, we figure out that when they want to follow the dominant eigenvalue which is the larger of the two numbers even if there both negative, we figured out that -5 is bigger the -10.3488

That means all the solutions in between are trying to approach this C1, this dominant axis, it is dominant because it has a larger eigenvalue and that is why all these red curves you see are approaching this dominant axis.3502

They might follow the other axis parallel for a little while but ultimately they all end up getting close to the dominant axis and asymptotically approaching it as they all dropped to 0.3521

That is the end of our lectures on real distinct eigenvalues, this is part of the systems of linear differential equations chapter, we are going to have a couple more lectures, one for complex eigenvalues.3534

We will get totally different pictures there and one for real repeated eigenvalues that is all part of the systems of differential equations chapters, I hope you will stick around for that.3547

This is all part of the differential equations lectures series here on www.educator.com. My name is Will Murray and I really want to thank you for watching, bye.3557

Hi and thanks for joining us here on www.educator.com, my name is Will Murray and I’m doing the differential equations lecture series.0000

We have already had a lecture on linear differential equation, this is the second lecture and we are going to be studying separable equations today.0007

I hope you will join us and let us jump in.0015

The idea of separable differential equations is that you are going to write Y′(x) as (dy)/(dx) and try to separate the variables in the following form.0022

You are going to try get all the y’s on one side of the equation and put (dy) over there and then you try to get all the x’s on the other side of the equation, put the (dx) over there.0032

That will not always work but if you can then it is called a separable differential equation where you can separate it into all the y’s on one side and all the x’s on the other side.0043

If you can do that then what you can do is integrate both sides and you will get some function of y on the left and some function of x on the right.0054

What you can do after that is solve for y(x), generally in to a few steps of algebra to solve for y(x) and you will get that y is equal to some function of x then you will be done.0067

The basic idea of separable differential equation, after we work couple of examples I think it will make a better sense.0083

There is a couple of notes I wanted to mention before we get started, one note is that some equation are both linear and separable.0089

We just had a lecture on linear differential equation and we had a completely different technique for solving that.0098

That was the one linear was the type of equation where you can write it as Y′ + p(y)=q and the we have a different technique for integrating factors for solving linear differential equations.0104

For some differential equations, they are both linear and separable so you can use the technique I just thought you or the linear technique to solve them.0119

If you do have a choice though, you probably want to go with the separation technique.0129

If you do have a choice the separation technique is usually a little bit less work than going through that whole integrating factor business, try to go with separation first, that is your first try.0134

Second very important note is remember that I said you are going to get something in terms of y × (dy)= something in terms of x × dx, and then you are going to integrate both sides.0147

What is important there is that you add the constant when you do the integration and not at any other step, it is very important that you do it exactly at that step of integration.0163

You do not have to add the constant to both sides you can just do it at one side because if you add the constant to both sides then you could just combine the 2 constants on the same side.0179

You only have to add the constant on one side but it is very important that you do it and it is important that you do it when you do the integration.0189

The reason it is so important is because after you do the integration there is usually a couple of steps of algebra for solving for y in terms of x.0197

A lot of times that will end up tangling up the constant into the equation and you have to let that happen.0207

You have to let that constant get tangled up into the solution, this is a real change from what you learned back in calculus 1 and 2.0213

Calculus 1 and 2 which you learned is whenever you do an indefinite integral, you just tack on a (+c) at the end.0226

You can do about 50 problems and then go back at the end and just put (+c) at the end of every single one.0232

Not so in differential equation, you got to be really careful with the constant and be careful when you insert it.0239

The time to add it on there is when you do the integration and every thing you do after that, you have to keep track of the constant as it gets tangled up into the equation.0246

Let us do a few examples and you will see how that constant plays out and how it really becomes part of the solution in a more significant way than it ever did in calculus 1 and 2.0255

For our first example here, we are going to find the general solution to the following differential equation Y′(x)=1/2 y(x).0269

Remember the whole point of separable differential equations is that you write the Y′ as (dy)/(dx) and then you try to separate the x’s on one side and the y’s on the other.0277

Let me write this as (dy)/(dx)=1/2 and that is y(x) that is not y multiply by x.0289

I will just write it as y and let me try to separate the (dy) and y on one side and then I’m going to multiply the (dx) over to the other side.0297

If I pull the y over the left hand side I will get dy/y on the left and move (dx) the other side is equal to ½(dx).0311

I have successfully separated my y’s on one side and my x’s on the other side, I can take the integral of both sides.0323

The integral of (dy)/y is just natural log or technically the absolute value of y.0332

The integral of ½(dx) is just 1/2x, now this is when I’m doing the integration and this is when I’m going to add a (+c).0339

I do not need a (c) on both sides, you do not need a (+c) here because you could just move it over and combine it with the other constant on the other side.0350

It is very important that you have a (+c) on one side or the other and that you incorporate it at this step.0365

Let me emphasize that, must add (c) when you do the integration.0374

Let me really emphasize that you go to do that when you do the integration.0392

The rest of it is just solving for y, in order to undo the natural log I’m going to raise (e) to the both sides, (e)natural log, that is the value of y is equal to,0401

Be careful, this is e (1/2x + c) is not, let me really emphasize this, I will this in red, this is not e 1/2x + c.0413

It is also not equal to e1/2x + ec, it is neither of those things.0429

It is really e(1/2x + c) and the way you resolve that is e1/2x × ec.0438

Ec is a constant so I’m going to call it a new value so I’m going to call it k and on the left hand side we get the absolute value of y is equal to just constant × e1/2x.0448

Y is equal to, technically + or – (ke)1/2x but that is still just a constant, I’m going to call that, what can I use for a new constant?0464

Y is equal to I will use a new constant A(e)1/2x.0483

Let me emphasize here that this is not like calculus 1 or 2 where you can just add on a constant at the end, we would have got in0494

Let me write this in red, we would have got in e1/2x and then we would just tack on a constant at the end.0505

It does not work like that because we really need to keep track of a constant and the fact that it turned into a multiplicative constant and not an additive constant.0514

We are finished with that problem but let me recap the steps here, I wrote Y′ as (dy)/(dx), I wrote y(x) as just y, that is not y multiplied by x.0528

The reason it says y of x here is to remember that y actually is a function of x, I wrote this as ½(y).0537

The whole game for separable differential equations is to get all the y’s on one side and all the x’s on the other side.0547

I moved this y over and I moved this (dx) over this side, so I get dy/y=1/2(dx), integral of dy/y is natural log of the absolute value of y.0555

Integral of 1/2 x is (½ x + c), it is very important that we add the (c) at this step because this is where we are doing the integration.0571

Then we solve for y, so we raise (e) to both sides and you have to be careful with your rules of exponents here e1/2x+c is = e1/2x × ec, not (+) ec.0581

We got the absolute value of y is equal to, ec is another constant and I will call that k.0597

Absolute value of y is = to (ke)1/2x , y just turns into + or – k which I call the new constant A,( Ae)1/2x.0603

I wanted to emphasize here that really the role of the constant, you have to add it as this point and keep track of it.0616

Keeping track of it show you that it is a multiplicative constant and not an additive constant, this would be wrong if you give it as an additive constant.0625

Wrong in the sense that it would not satisfy the differential equation, it would not make the differential equation true anymore.0635

That is why you really have to be careful with that.0641

Let us try another one here, we have to find the general solution to the following initial value problem.0646

We have (yY)′ + x =0, y(0)=3, remember the goal with the separable differential equations is you write your Y′ as dy/dx and we are going to write y × dy/dx + x=0.0655

The idea is to try to separate the variables, get all the x’s on one side and get all the y’s on the other.0677

In this case we get y × dy/dx, I will that x on the other side and get –x and then if I multiply dx by both sides I will get y/dy=-x/dx.0682

Now I have successfully separated my variables x’s on one side, y’s on the other.0700

Take the integral of both of these, the integral of y/dy is = y2/2.0707

The integral of –x/dx is –x2/2, but I just did my integral so I have to put the constant right here.0713

I’m going to try to simplify this a bit, I’m going to multiply both sides by 2 so I get y2 is = -x2 + now 2c is just another constant so I’m going to call it (k).0725

Add the x2 + y2= k and now I’m going to use my initial condition in order to find the value of that constant.0746

I’m going to plug in x=0, y=3, I plug those in y(0)=3, that tells me 02 + 32=k, that tells me k=9.0759

If I plug that back into my general solution there, what I get is x2 + y2=9.0781

That solves my initial value problem, let me remind you of each of the steps that we went through there.0801

We started out with by converting this Y′ into dy/dx and then I was trying to get my x’s on one side and y’s on the other.0806

I got my x on the right, I still have dy/dx on the left so I multiply that dx over to the other side and get x/dx on the right.0816

Now I integrate both sides, integral of y/dy is y2/2, integral of x/dx or –x/dx is –x2/2 + a constant.0827

You do not have to add the constant on both sides but you do have to add it on one side, it is very important that you do it right here when you do the integration.0837

I was going to solve for, try to simplify as best I can, to get rid of those 2 I multiply both sides by 2, get a 2c over here which is 2 × the constant, which is another constant so I called it k.0847

In order to find the value of k, I used my initial condition x=0, y=3, I plug that in right here and I found the value of k is 9.0862

I substitute that back in to the general solution and I get x2 + y2=9.0874

On example 3, I’m given a differential equation Y′(x) + xy=x3, I have to determine if this thing is separable.0885

The way I’m going to do that is I’m going to write my Y′ as dy/dx + xy=x3.0896

I will be wanting to multiply my dx by both side so I’m trying to move and keep my dy/dx on one side and move x3 - xy over the other side.0909

What I would like to the is to factor the right hand side, I want to factor this into something of the form of a function of x × a function of y.0924

I can not factor that into that form and because I can not factor that, this equation is not separable.0948

We can not do this, we can not factor that in that form.0957

This equation is unfortunately is not separable, we can not use our technique of separating the variables to solve this differential equation, it is not separable.0973

That is all really we are asked to find for this example, we are asked to check whether this differential equation is separable.0993

However I will add a note here, in a parenthesis, (However, it is in linear form.)0999

I will remind you what that form is, the linear form for differential equations is Y′ + p(x)/y=q(x).1012

This is in linear form so you could solve this differential equation using the technique we have learned on the previous lecture on linear differential equations.1026

It could be solved using our technique for linear differential equations.1038

That is something that I have a separate lecture here in the differential equation lecture series for linear differential equations.1053

There is another lecture here in the differential equation lecture series for linear differential equations where we learned how to solve this.1073

We used something called the integrating factor to solve the linear differential equation.1081

If you want you can just check that lecture and you can see exactly how to solve this differential equation using that technique even though our separation of variables technique did not work on this one.1086

Let me just remind you what we did here, we started out by writing Y′ as dy/dx, that is a quick first step for all separable differential equation.1099

You put everything over the other side and you try to factor it into something that will easily separate the x’s and y’s.1110

Unfortunately this one, you can not factor it in that form so this differential equation is not separable.1120

However, that is essentially the end of the road as far as separation techniques go but I did notice this original differential equation is in linear form Y′ + p(x)/y=q(x).1126

That was our linear differential equation form so you could use the linear differential equation technique that we learned in the previous lecture to solve this one.1144

You would not give up hope completely on this one, you just have to watch a different lecture to learn how to do it.1154

In our next example, we want to solve the initial value problem 3x – 6y × square root(x2 +1), dy/dx=0 and we also have the initial value of y(0)=4.1161

Again, let us try to separate the variables there, I think I’m going to write 3x= 6y × square root(x2 + 1) dy/dx.1177

I can divide both sides by 3, that will give me 1 here and 2 here, what else I’m going to do?1195

I think I’m going to move the x’s over to the other side, keep the y’s over on the right.1205

I get x/square root(x2 +1) = on the right I’m going to move the dx over to the other side as well, so dx here.1209

On the right I’m going to get 2y/dy, and now I have successfully separated all my x’s to the left and all my y’s to the right, I can integrate both sides.1223

2 separate integrals, one in terms of x and one in terms of y.1237

The y one is definitely going to be very easy that is just y2.1242

The x one looks more difficult, I’m going to do (u) substitution there.1246

I’m going to use u=x2 + 1 and du=2x/dx and if I move that 2 over the other side, I’m going to have 1/2du=x/dx which is what I have in the integral.1251

Let me convert that integral, I see I have 1/2du/square root(u), if I pull that half out to the outside I got ½.1273

The square root of u that is u-1/2du, this is all still equal to y2 and I need to integrate that.1287

Let me remind you the power rule, it says the integral of un/du=un+1/n+1 + the constant.1297

Here my n is -1/2, n + 1 is ½ so I get u1/2/1/2, that is n + 1.1309

I still ½ on the outside so ½ on the outside multiplied by that and I have to add my (+c) right now because now is when I’m doing the integration.1322

This is equal to y2, let us keep going up here, those half cancel each other out and I get u1/2.1337

What was u? u was x2 + 1, so I get the square root of x2 + 1 + a constant is = to y2.1346

Let me just switch my sides here, y2= square root of (x2) + 1 + a constant.1359

I’m ready to use my initial condition to find my constant, now this initial condition is telling me x=0, y=4, I plug those in here.1367

Y2 will be 42=square root of 02 + 1 + (c).1379

I get 16=1 + (c) and so my c =15, I plug that back in here I will get y2=square root of x2 + 1 + 15.1387

If I solve for y, I see that my y is positive here so I’m going to take the +square root, y =square root of all of that.1411

The square root of x2 + 1 + 15 and then the square root of all of that and now I have solved for my y.1421

Let us recap what happened here, I was trying to separate my variables so the first thing was I moved all of this stuff over to the other side.1442

That is how I got 6y root x2 + 1, dy/dx on the other side.1452

I have 3 on one side and 6 on the other, it seems like I should cancel there.1458

If I divide both sides by 3 I get a 2 over on the right, that is where that 2 comes from.1463

I moved the x2 + 1, I divided that over to the denominator and I multiplied that dx over the numerator.1469

That just leaves me with 2y/dy on the right and x2 + 1 on the left.1476

The key point there is I have separated all x’s on the left, all y’s on the right and now I can take the integral of both sides.1484

The integral of 2y is just y2, I’m not worried about the constant there because I know I’m going to add it on the other side as I do my integral for x.1493

That integral turns out to be a little more complicated so I did a substitution u=x + 1, du=2x/dx and so x/dx is ½ du.1502

I know it is that because I have a x/dx there, that turned into ½ du and then I have square root of u in the bottom which is the same as u-1/2.1513

If I integrate that using the power rule back from calculus 1, I raise it by 1, -1/2 + 1 is +1/2.1529

I have to divide by ½ but that cancels out what my ½ out here, that is ½ and that is cancelled out.1540

This is when I do the integration, this is when I add the constant, it is not ok just add the constant at the end in differential equation, you got to add the constant when you do the integral.1549

Add (+c) when you do the integral.1561

The reason you have to do it when you do the integral and not just tack it on the end is because there is several more steps of algebra coming.1575

That (c) gets tangled up in to the equation, in particular it got put under a square root here.1583

You really have to keep track of the (c) as you go through all the algebra afterwards.1590

That is what I was doing here is I was trying to solve for y2 and I was using my initial condition, plugging in y= 4, plugging in x=0 to solve for c.1594

I solved for (c) and I got (c)=15 but then I still have to solve for y.1608

I took the square root of both sides and the (c) goes under the square root, that 15 goes under the square root, I have to keep track of it from now on.1614

I could not just tack it on at the end, I really have to keep track of it and the role it plays in the equation.1623

My final answer there is the square root of x2 + 1 + 15.1630

A little bit complicated there but the basic technique is not too bad.1636

We got one more example here, example 5 is to solve the initial value problem Y′(x) =x2y and y(0)=7.1644

My Y′ there since we are going to treat this as a separable differential equation, I’m going to write that as dy/dx = x2y.1656

I’m not going to worry about that initial condition yet, that will come in at the very end is when you actually use the initial condition to solve for whatever arbitrary constant you find.1667

In the meantime what I’m going to try to do is try to separate all the y’s on one side and all the x’s on the other.1677

That is just a matter of multiplying and dividing, if I divide both sides by y, I will get dy/y on the left, multiply both side by dx I get x2 dx on the right.1684

I have successfully separated the y’s on the left and x’s on the right, I can integrate both sides.1696

The integral of dy/y is the natural log of absolute value of y.1703

The integral of x2 is x3/3.1708

I just did my integration I got to add on a constant right now, that is very important that you add that constant right when you do the integration, not earlier and definitely not later at the end.1713

I’m going to try to solve for y, I’m going to raise e to both sides, e to the natural log of absolute value of y= be very careful here, e to all of the right hand side x3/3 + c.1725

That all gets put in the exponent of e and not just the x3/3 part, I got absolute value of y is equal to1738

My laws of exponents here tell me that this is e to the x3/3 × e to a constant, not added to e to a constant but multiply by e to a constant.1748

Y is equal to (+ or -) e to a constant, think of that as a single constant e to the x3/3.1759

If I think of that as a single constant, I’m going to write that as k × e to the x3/3 because e to a constant is still a constant and (+ or -) a constant is still a constant.1771

I got y = ke to the x3/3.1784

That is the general solution for my differential equation, it is as much as I can derive just using the differential equation.1788

However, I also got the initial condition, I have been given that and so I’m going to use that to go a little bit further and figure out the particular value of that constant.1799

If we have not been given that initial condition we will just stop right now with the general solution but since we have that initial condition y(0)=7.1810

I plug in y=7 and x=0, 7 for y and 0 for x, 03/3, in the 0 that is just 1, that is equal to k.1821

I get 7=k and if I plug that back into my general solution, now that I know what k should be, I get y is equal to 7e to the x3/3.1836

That is my particular solution to the initial value problem, it is a solution to the differential equation and to the initial condition that we are given.1855

Let me emphasize here that it was really a key that I added the (+c) when I did the integration.1866

I could not do it at the end at this step right here at the end like you would in a calculus 1 or 2 problem.1874

In calculus 1, you would just tack on a (+c) at the end and everything is fine.1881

Here you got to add the (+c) right when you do the integration and then what happens is that (c) gets tangled up in the equation after that.1885

Let me go through those steps again just to make sure that they all make sense to you. I wrote Y′ as dy/dx -- that is a very common technique for separable differential equations.1895

That is still equal to x2y, I’m going to cross multiply and cross divide to get all my x’s on the right and my y’s on the left.1907

If I divided that y over to the other side and I multiply dx over the other side, I get this nice separated form with dy/y and x2 × dx.1922

I got all x’s on one side and all y’s on the other I can integrate both sides.1934

Integral of dy/y is natural log of y, the integral of x2 is x3/3 + c.1939

You got to add that (c) when we integrate and then all the steps after that you got to keep track of what is happening to the (c), you can not just tack that (c) on at the end.1947

To undo that natural log I raised e to both sides that gave me just absolute value of y on the left.1958

On the right it gave e to the (x3/3 +c), now you can not write that as e to the x3/3 + c, that is not the same thing, that would be very bad.1968

It is also not equal to e to the x3/3 + ec, that is also bad.1982

We are using the laws of exponents here, xa + b=xa × xb.1988

This is e to the x3/3 × ec, I pull that over to the other side.1998

Because I have an absolute value here, I just made a (+ or -) but that whole thing (+ or -) ec is just on e big constant that I called (k).2005

In order to find the value of K, I used the initial condition, this is telling me right here that x=0, y=7 .2013

I plug those values into this general solution here y=7, x=0 and e0 is just 1 and this just turns in to k =7.2026

I figure out that my k is 7, plug that back into the general solution and I get my specific solution y=7e to the x3/3.2039

That wraps up our lecture on separable differential equations, I hope you have enjoyed it as much as I have.2050

We got a lot of other lectures available on a lot of other topics on differential equations.2058

This is the very first topic, we got all kinds of stuff available on systems of differential equations, on solving differential equations by series.2063

We got stuff on second order differential equations, all the different solution techniques that you are going to learn in any differential equation course.2073

We got some stuff on numerical solutions to differential equations, we got partial differential equations in 4a series.2082

They are all here in the differential equations lecture series on www.educator.com.2089

I hope you will stick around and join me for all those other lectures and enjoy learning differential equations with me.2097

My name is Will Murray and this is the differential equation lecture series here on www.educator.com.2104

Thanks for being with us, bye.2109

Hi and welcome back to differential equations lectures here on educator.com.0000

My name is Will Murray and today we are going to be studying systems of differential equations, where the matrix that gives the coefficients for the system turns out to have complex eigenvalues. 0004

So we already have a lecture on systems of differential equations, we already saw the basic idea where you find the eigenvalues and eigenvectors the matrix and then drop those into a solution.0015

So if you have not watch that yet we are probably be good to go back and work through that lecture before we learn about the case on complex eigenvalues because this 1 is a little more complicated.0026

So on this case in this lecture I am going to kind of assume that you have already done the 1 with real eigenvalues and we will move on more complicated case of complex eigenvalues.0036

So let us see how that works out.0046

Let me just remind you that to solve the system of linear differential equations X′, this is vector x here and the vector X′ and then this A stands for a matrix.0048

Usually we will solve 2 matrices.0061

The way to solve that to find the eigenvalues and the eigenvectors of the matrix A so that sort of the first step in solving these things.0064

And if the eigenvalues are complex then they will occur in conjugate pairs, meaning you get if A + BI is 1 eigenvalue then A- BI is the other eigenvalue.0073

And that really comes out of the quadratic formula remember the quadratic formula tell you that R = -B + or - the ² root of the ² -4 AC and so all over 2A. 0086

And so if you get a - number under the ² root then that is the case where you are going to have complex numbers as your eigenvalues.0100

And so if you got a - number of the ² root then you are really going to get -B + or - that complex value.0108

So to tell you always get these things in conjugate pairs A+ BI and A- BI.0115

So here is how you handle something like that.0122

You choose 1 of the eigenvalues I always choose R = A + BI just because it is 1 less - sin to keep track off.0123

And if you find corresponding eigenvector B, that can be a little challenging to find the eigenvector when using complex numbers but will do some examples here and you will see how it works out.0134

In any form the complex solution it still the same form as before remember our old form was in the RT where R was the eigenvalue x the eigenvector B.0143

So we are doing the same thing here we are forming the complex solution where we put in the eigenvalue as the R.0153

A + BI and those we multiply by the eigenvector B.0160

But the problem is that still has complex numbers in it so let me show you how you can expand that out and get to real solutions.0164

So what you do is you use Euler formula which is E I θ = cosine θ + I sin θ.0171

So to remind you what we had on the previous page we had an E A + BI that was the R x T & we want to expand that out into E AT x E B IT.0187

And the E VAT is no problem we are just gonna leave that the way it is if the B E IT that we are worried about.0210

We think of this as this B E IT B IT as EI x BT and then , we use Euler formula on that so it expands out the BT is playing the role of the θ here.0217

So we have E I x BT and that expands out into cosine θ + I sin θ that cosine of BT + I x sin of BT.0231

So you are gonna expand out that in the B IT part into a cosine BT + I sin of BT and you still have all this time some eigenvector which will itself have complex numbers in it.0242

And so you multiply these complex terms the cosine BT and the i sin of BT you multiply those in eigenvector and so you are going to get some complex numbers inside the eigenvector.0255

Then, what you do is you separate that eigenvector with the E terms inside it into a real part and imaginary part. You are going to have some terms which just a real numbers and some terms which have imaginary have I multiplied by some real numbers.0270

In what you do is you take each 1 of those and you make those into your 2 solutions and you can drop the I now.0287

This first 1 becomes your X1 solution and the second one without the i becomes X2 solution and then you attach constants to each one a C1 and C2 and that becomes your general solution in terms of real numbers.0295

There is no more I after this because you separated out the real part and the imaginary part and then you drop the eye off the imaginary part.0315

We will see how that works out I think I make more sense after reducing examples but that is some kind of the general overview there and let us see how that actually plays out with some specific problems.0323

Let me tell you that we are going to graph these, so in order to grap the solutions the trick here is to ignore the EVA T factor at first.0335

What you have if you ignore that factor is a bunch of terms which have things like cosine of BT and sin of BT and the trick is to pick values of T that are going to make BT equal to some nice common values that it is easy to plug into cosine and sin.0347

Ones I like our 0 π over to π 3 π over 2 and 2 π so for example if B were equal to 3 then what you have is cosine of 3T & the sin of 3T.0368

And so you want to pick values of T that will make 3T a multiple of π over 2 so, I would pick T equal to 0 and then the 3T would be 0 and I probably pick T equal to π over 6 and the point of that is that 3T would be π over 2.0388

So that is an easy value when you put π over 2, when assigning cosine, and it is very easy answers.0409

And probably take pick 3 equal to π over 3 because then, 3T would be π and then T it would be I would take π over 6 π over 3 π over 2 because then 3T would be 3pie over 2.0415

Finally I would take T = 2 π over 3 so I am going by multiples of π over 6 because π over 6 x 3 is π over 2.0439

Finally I would get 3T would be 2 π.0449

Appointed that is when I saw B was equal to 3, I picked values of T that when I multiplied by 3 are going to give me 0 π over 2PIE 3PIE over 2 or 2PIE.0454

I am going to graph the points that arise very easily.0467

What is going to happen in all of these is you are going to get ellipsis.0473

When you graph out these values of T you are going to get ellipsis.0475

Your basic graph is going to have ellipsis and then you are going to look at this factor of E AT and essentially if A is positive, that means E AT is getting bigger and bigger and bigger so you are gonna draw this ellipsis getting bigger and bigger and bigger.0478

It is going to be spiralling outwards.0496

If a is less than 0 than E AT would go to 0 which means you are going to draw this ellipsis restricting down smaller and smaller and so you can have a pattern that spirals in origin like this. 0499

In the A positive case you have a pattern that gets bigger and bigger and so you have like ellipsis but their spiralling bigger and bigger so will see that as we get some examples.0517

Just like all the other problems if you are given initial conditions and you can use those to solve the constant C1 and C2.0529

It is deftly time that we got some examples and you try this out with some actual numbers so worked the first example together.0536

We got X′ = 31 -21 and x X so remember all of the systems problems start out the same way always start by finding the eigenvalues of the matrix and when you do that if you subtract R down the main diagonal.0546

I got 3 - R x 1 - R so 3 - R x 1 - R - -2×1 that is +2 = 0 so if I multiply this out I got R ² - R - 3Rs that is - 4R +3 +2 so we got +5 = 0.0563

That is not a quadratic equation that factors nicely so I am going to go straight quadratic formula. 0591

R = - B so positive 4 + or - B ² -4AC B ² is 16 4AC is 4×1×5 so -20 all over to A which is 2 and so that is 4 + or - .0597

Now 16-20 is - 4 so I got the ² root of -4 so that is 2 I all over 2 which is 2 + or - I .0617

So remember, I always I said these roots of the characteristic equation the eigenvalues they always occurring conjugate pairs always have A + BI and A - BI so here we got 2 + I and 2 -I and the nice thing about these is you really only have to work with 1 of them.0627

So we are just going to pick R = 2 + I and will play that back in matrix so 3 - if I put 2+ I in for R that is 3-2 is 1 - I and also have 1 over here -2 over here 1-2+I is -1 - I and then, I want to set this up as to find the eigenvector.0646

So I will find a vectors such that when you multiply that by that vector you get 0 and little trick those works for me is to plug-in Y = 1 and will figure out what the X should be.0682

So let us this trick it is very helpful for finding eigenvectors and I think it is more useful if we look at the bottom row here0694

So I got is -2 x X-1 - I x 1 = 0 and I am gonna solve for X and I see that I got X if I move the -1 -I over other side I get + I over -2 so that is -1/2-1/2 I so my eigenvector would be -1/2-1/2 I A on top and 1 on the bottom.0705

I am really not thrilled with having all those - sins and with having those fractions.0744

Remember you can multiply eigenvector by a scalar if you want to make a little cleaner so that is what i am going to do here so i am going to multiply by -2 and I am going to get that gives me 1 + I in the and on the numerator in the first quarter of the vector and -2 in the second core in a second coordinate.0747

so now I've got my eigenvalue and my eigenvector and I am going to plug that into my generic solution which is number generic solution is E in the RT were R is the eigenvalue x the eigenvector.0766

so E in the RT the R was 2 + I used 2 + IT x the eigenvector is 1 + I20782

I am going to separate out the E 2T so, I will just remind you that E to the 2+ I T = E 2T x E IT and so all right E 2T separately and that is not going to affect the immediate future here.0795

And now E IT I am going to use Euler formula remember, Ei θ = cosine θ + I sin θ so E 2T as our E IT so I had to deal with here that some cosine T + I sin T.0818

And I am still multiplying that by the eigenvector 1 + I and 2 and I am going to multiply that in eigenvector so, E 2T outside so if I multiply that in to each coordinate the eigenvector gets a little bit messy.0845

I get cosine T + I sin T I am going to multiplied it by 1+ I cosine T multiplied the first term by I now I x I so I got -1 sin T and the bottom I get 2 cosine T +2 I sin T 0861

Now I said that the goal here is to separate this out into a real part and imaginary parts so, each 1 will have the E 2T in the 2T if I just collect all in real terms over here and I will collect my imaginary terms over here and we will see what we get for each 1.0887

For the real terms I see at the top I have a cosine T - sin T bottom I got 2 cosine T and now , the imaginary terms as you have it I x the sin T + a cosine T so feel that in sin T + a cosine T and 2 sin T on the bottom.0908

So, those are my real and imaginary part of the solution remember I said you can take each 1 of those and make those into separate solutions that is going to be our first solution our X1 and that is can be our second solution X2 .0933

And then just take each 1 of those attaches a constant to each 1 and you can write down the general solution right away.0953

So, our general solution is just C1 x that first solution C1 x E 2T x cosine T - sin T and something is bothering me a little that I see what is bothering me I have a -2 here and I forgot to bring that -2 along that - sin along for the ride so they fill that up here.0962

There is - here - that should be - and this should be - so all my 2s through there should be - so we go ahead and fix that.0994

-2 cosine T sorry about that little mistake there + C 2 E 2T x the other solution sin T + cosine T all over -2 sin T .1006

And that is our general solution to that system of equations or to that matrix differential equation, if you prefer that for so let's start recap how we got that.1027

First thing for any of these problems is to find the eigenvalues and eigenvectors the matrix that is related here.1047

Subtracted are the main diagonal, I took the determinant , I cross multiplied to get the determinant, I got this quadratic equation for R which did not factor nicely so I had to use the quadratic formula and simplify down to 2 + or - I. 1056

Remember these complex 1s you only have to take the first route .1069

I always was to go for though the 1 with + just to make me have to do with fewer - sins later so I used R = 2+ I plugged that back in the matrix for this R here I got 2+I in there as I gotten 1 - I and -1 - I .1074

On the main diagonal here and try to find the eigenvector here so putting a Y 1 for the Y and I try to figure out what action be correspondingly so looking at the second equation here that is -2 X -1 -I =0 .1090

And I solve that out for X plugged that back into my eigenvector and I see I got some nasty fractions but member it is okay to multiply eigenvectors by scalars if necessary to clear some nasty fractions . 1107

Since of this -2 just multiply -2 in top and the bottom and so, I get a new eigenvector which does not have any bad fractions anymore and out with his in vector I plug at back into my generic form for the solution E of the eigenvalue x T x the eigenvector.1123

In E + 2 +E IT I expanded that out in E 2T and in the E IT just say the E 2T on the sides not really get into anything more .1148

The E IT expands out by Euler formula that was Euler formula up there so expands out into cosine T + I sin T .1156

And then we multiply that in eigenvector which requires keeping track of all the complex terms, it is a little messy in the top here we multiplied it through the -2 bottom and then we look through all those terms segregate out the real terms and the imaginary terms.1162

So here, I've segregated all the real terms and here I segregated away all the imaginary terms and if I factored out and I from all the imaginary terms is all the terms it had and I affected out group them altogether.1184

And it turns out that each 1 of those gives you a fundamental solution to your original system so you do not have to consider the I anymore you just take each 1 of these and you form a solution to original system .1202

Since I d1 that here I took just copied those 2 solutions down C1 and C2 onto each 1 and then put them together to get my general solution. 1219

So you want to hang on this general solution we we're d1 with all the competitions what we will be working in the next example is taking a look at the graph of this and plug in some values for T and see how it plays out on some actual axis.1228

So in the second example, here we are going to graph the solution trajectories the previous system of equations .1243

I got here is the solution that we derived in the previous example so, if you are not sure where this equation came from then you might want to go back and rewatch example 1 because it was example 1 were we derived the solution 1257

What really doing now in example 2 is working with this equation and trying to graphs and solution trajectories so, let me show you how that plays out the setup some axis here.1272

The nice scale so 1 2 on my X 1 axis here's X2 1 2 there .1284

So remember the trick here is you do not worry about the E the 2T term at the beginning so let me not worry about in E 2T .1307

What I am going to do is just going to look at this part to start with just a graph 1 of the solutions because I am going to take the thing that is taken C1 = 1 and C2 = 0 and I am also going to ignore the E2T at the beginning .1315

We will see how that what role that plays in a moment so, I am gonna plug in some values of T into that solution and I am going to see what kinds of numbers I get 1333

So, I am going to take T = 0 I want to take the values that are and can have me plugging in multiples of π over 2 so T is equal π over to T = 3 π over 2 and T = 2 π .1344

So, think that will be enough values of T to get an idea of what the shape is the plug-in T 0 plug-in and here we get cosine of 0 that is 1 - sin of T that 0 and - 2cosine T that is -2 so that gives me 1 -2 .1367

Basically I am just looking at the cosine terms drop out if I plug in T = π over 2, I see well, now for π cosine is 0 in the sin is 1 so going to get the - 1 on the top and 0 on the bottom .1385

For T = π the cosine is -1 and the sin is 0 so I get - 1 on top and then -2 x -1 so, positive 2 on the bottom 3 π / T cosine a 0 sin is -1 so I got positive 1 and 0 and finally T = 2 π .1403

Remember, I am plugging them all into this first solution right here T= 2 π is the same as T = 0 so I get positive 1 on top from the cosine and -2 on the bottom .1423

Slowly plot those out when T = 0 I see I got 1 -2 so 1 on the horizontal axis -2 on the vertical there is T = 0 and T = π over 2 got -1 0 -1 0's there is π over 2 and T = π I see I have -1 2's that goes up here there is π and T = 3 π over 2 I see 1 0 so there is 3 π over2 and T = 2 π is back to 1 -2 .1439

So that is back the same place we started, it is T = 0 and if I connect these up it is pretty clear that I have an ellipse itself is a circling around forever but, it is it is not a circle it is an ellipse so it is the ellipse in around forever .1479

And we know it is going in that direction in the clockwise direction so that is that sort of the basic shape of the graph .1501

However there was something I was not considering yet which was that there is a E2T in there as well .1513

So, what happens is, the E2T course that goes to infinity so what that does is it sort of makes this ellipse get bigger and bigger and bigger each time you circle around.1521

So, this makes it spiral outwards because basically that is because the eigenvalue 2 is positive there so what I am going to do is say this blue ellipse is some sort of a first draft that is not the actual graph.1533

I will make the actual graph will make it red and what I want to do is have that ellipse Getting bigger and bigger as it goes around and around.1560

I will draw few more trajectories here basically, they're all ellipsis Except that they are getting bigger and bigger every time they go around.1576

And even down here in the origin they're getting bigger and bigger following the same general shape is that ellipse .1585

But getting bigger and bigger each time so really the blue graph here is not the final graph that is not the answer, it is these red spirals that are the true shape of the solution trajectories.1605

So, let me show you how we derived that I remember we only need to look at 1 essentially if you look at the other at the other solution here, you are just going to end up with the same points just all shifted around by π over 2 .1620

So you would end up drawing the same thing if you looked the other 1 so we just looked at the first 1 the cosine T - sin T on top and then -2 cosine T on the bottom .1635

And what we did was we plug-in select values of T to give us a shape of the graph C = 0 π over 2 π 3 π over 2 and 2 pies.1645

We plugged those in and we got different points year 1 -2 -1 0 -1 2 1 0 and went back start 1 -2 .1656

So, I grasped each 1 of those points is what I did here and I got an ellipse and I also got the direction of travel of the ellipse so just by graphing those is how I got this blue ellipse .1666

Kind of going around clockwise but then I noticed that I also had this term of E 2T and that is going to be magnifying the size of the ellipse each time it travels around .1682

So a set of a pure ellipse, what I actually have is an elliptical spiral that is getting bigger and bigger each time he travels around . 1694

So that is what you see here with the red curves the red curves are actually the 1 that are actually giving me the solution to the problem .1702

So for example 3 were to find the general solution following system .1712

3 -5 5 -5 so, just like all the others here the first very first step in all of these is to find the eigenvalues and the eigenvectors of the matrix.1719

So, remember the way you do that is you subtract R from the main diagonal and then you get yourself a quadratic equation for R, 2901, were to cross multiply here so got 3 - R x -5 - R -5 x -5 so +25 = 0+25 = 0 and let's see how that plays out I see them and have a positive R ² +5 R -3 R so +2 R and now got -15+25 so that is +10 = 0 .1731

Again that is not a polynomial that factors nicely so in order to solve that I have to use a quadratic equation R = =B 2 + or - now be ² is 4 -4 AC , AC is 1×10 so -40 all over 2A so that is - 2 + or - 4-40 is -36 so I am going to get complex solutions the ² root of -36 is 6I still divided by 2 so I get -1+ or -3 I .1780

So again I get 2complex roots remember they always occurring conjugate pairs -1+3 I and -1 - 3I 1822

The nice thing about solving systems with complex eigenvalues is you only ever have to treat 1 of the of the eigenvalues so, we are going to take the 1 with the R -1+3, I re-plug that back in matrix for that R on the main diagonal so what we will get is 3 - R 3 - -1 is 4 -3 I -5 and 5 and -5 - -1 is -5+1 it is -4 -3 I .1834

And again , we want to find eigenvector corresponding to that eigenvalues set th= 0 , and as usual I am going to put a 1 in here and I am in a try to figure out what action be to make it work and I am going to look at the second equation that is often useful in these complex in these matrices with complex eigenvalues .1875

So looking at 5X -4-3 I = 0 using that second equation there so if I solve for X I move the 4+3 I over the other side X = 4+3 I over 5 .1896

So my Eigen vector is then read that is 4/5 + 3I +3/5I on top and 1 on the bottom .1921

Again I do not like these fractions and because it is an eigenvector it is okay to multiply the whole thing by a scaler in order to eliminate whatever fractions you might want to get rid off.1932

So to multiply this whole thing by 5 and I will get on the top there i will get 4+3 I and on the bottom of this get 5 so it is a much nicer eigenvector so now have an eigenvalue and an eigenvector .1941

So I am going to plugged this and I got remember my generic formula is in E RT x V where R is the eigenvalue and V is the eigenvector. 1948

So in this case my Eigen value is -1+3 I all of that x T and now my eigenvector is 4+3 I / 5 1970

The point is I like to sort that out into real parts and imaginary parts, I may be using Euler formula and but First I am going to right E -1+3 I see the separate that out as E -1T x 3T .1989

Of course that should have been an I so E -1 is E 3 IT remember this Euler formula was E I θ = cosine θ + I sin θ .2009

And the key part here is that , θ in question is multiplied by I that is going to be 3T in our case that is our θ right there the 3 and the T right there .2029

So let me expand this out this is E -T x now cosine θ + I sin θ so that is cosine of 3T + I sin of 3T x my eigenvector which is 4+3 I x 5.2043

Now , I will multiply the real and complex parts the real and imaginary parts in eigenvector so i am just going to preserve this in the - T it is not really doing anything for the near future.2065

Multiply everything in eigenvector so I get 4 cosine 3T + 4 I sin of 3T is + 4 I sin of 3T +3 I cosine of 3T +3 I x I which is -1 so - 3 sin of 3T .2078

So that is the top part is a hard 1 in the room bottom I get 5 cosine 3T +5 I sin of 3T .2109

So, if I expand that out into a real part and imaginary part of E the - T x both of them in real terms over here + I E - T and all collect all my imaginary terms over here wasn't E - 2 in my T got run together a little bit there that looks little better .2121

So let me collect all the real terms on the left so 4 cosine of 3T and I see I have a - 3 sin of 3T in the bottom for the real side I have five cosine 3T .2147

Now , on the compliment on the imaginary side, Ii am going to collect all the terms that have I in them remember with the eye outsides would only to write an I inside 4 sin of 3T and 3 I cosine of 3 T I outsides up at 3 cosine of 3T .2165

On the bottom I see I have a 5 sin of 3T because I can go outside.2187

So remember, the way this works once you find your real solution in your imaginary solution you can click the I off the imaginary solution and then each 1 of those is going to be 1 of the terms of your general solutions .2193

That is X1 and that is X2, so you just put those together with constants to form your general solution.2209

Slowly go to do that X is C1 x E to the - T x that first solution 4 cosine of 3T -3 sin of 3T and 5 cosine of 3T + C2 x that second solution which was E - T .2217

Notice I am not writing the I anymore you do not need the I resolve this into real solutions 4 sin of 3T +3 cosine of 3T and then five sin of 3T on the bottom .2245

So that entire thing is our general solution we got 2 specific solutions but constants multiplied by each 1 and put them together to get your general solution.2264

Since the end of that problem in the next example , were to go through it and working to graph this see how this looks on an actual set of axis but before we do that I like to recap how we got there because the next problem were just in a start with this general solution will say where it is coming from.2279

So what we started out by doing was to find the eigenvalues and eigenvectors , the matrix the system from always E way, you always find eigenvalues and eigenvectors.2297

So that means, subtracting R from the main diagonal of the matrix cross multiplying cross those and subtract them which is how I got this quadratic equation, subtracting -5×5 some subtracting it turns into a positive 25 .2307

Simplify that and run the quadratic formula on it to get my 2 eigenvectors or much eigenvalues which are always conjugates of each other so they always look like a + or - B I so -1+ or + or - 3 I .2326

I pick the positive 1 -1+3, I plugged it back in for the R so plug-in -1+3 I in the R in the matrix as I got that matrix there and I want to find the eigenvector and my trick for doing that is always a start with the 1 in the y-coordinate.2341

Then you had to figure out what the X should be and it is usually useful to use the bottom line here so this is multiplying the bottom line x that vector 5X -4-3 I = 0 .2359

Solve for X which is pretty easy, plug it back in for the X values the eigenvector many notice all got these bad fractions like to remove the denominators but since it is an eigenvector can multiplied by a scalar and you can clear the denominators away.2372

Since I did hear multiplied by 5 and I got the eigenvector 4 +3 I over 5 .2389

So then, I plugged that back into my generic solution E RT x V so there is the eigenvector and there is the eigenvalue right there .2395

I expanded that into E - T which does not really do anything for a while and E3 IT which I used Euler formula up here to expand that out into cosine of 3T plus I sin of 3T.2406

And I multiplied that back in eigenvector so , we multiplied this in top and bottom here we got 4 terms in the top because it is a binomial x a binomial 2 terms in the bottom and then we segregate out the real parts of the solution and the imaginary parts .2424

Imaginary parts of the 1s multiplied by I, so we segregated those out we sorted them out into everything real and everything imaginary.2446

The imaginary we factor out and I and then each 1 of those becomes 1 of our solutions that were going to use for general solution so, that is the real 1 over there at the imaginary 1 over there except that we do not need an I anymore we just attach arbitrary constants C1 and C2 to it.2454

So that is how we derived that solution in the next example you want to hang on solution because we’re going to start with this equation and were going to go graph were going to see what it looks like .2473

So, let us go ahead and see how that 1 plays out so next example this is the exact I copy down the general solution from before but in bother writing down the C1 and C2 .2484

So, got E - T x that first vector E - T x that second vector .2496

If you just tuning in if you have lost the previous example for while, these 2 vectors they came out of the answers from the previous example 2506

So, you are wondering where these came from what we did in example 3 was we solve the matrix we found the eigenvalues and eigenvectors and then we went through some algebra to resolve the solutions in to these 2 pieces.2511

So if you have a just watch example might go back and check it out and see where the solutions come from .2529

Example 4 what were going to be doing is graphing it so I am a set up some axis here vertical axis here very vertical to see if I can know little better than that.2535

Little hard to draw vertical lines on these tablets but I will do the best I can do is a little bit better alright and looks like I am be going up to five some scale on there.2553

12345, now the key thing here is you want to be plugging in values of T that are going to give you multiples of π over 2 .2571

So , I am seeing in my solutions here I got a bunch of 3 T so far I want to get 3T to be a multiple of π over 2 then I see I got a make T is a multiple of π over 6 .2595

So, I am going to use here multiples of π/6 and so that will give me nice multiples of π/2 in the sin cosines which will be very easy values to plug-in and find nice numbers for.2610

So, multiples of π over 6 start out with 0 C = π over 6 T = 3 π over 6 which is the same as π over 3 and finally T = 3 π over 6 which is the same as to π over 2 and finally T = 4 π over 6 which is the same as 4 π over 3 .2625

And the point of doing that as a something I mentioned briefly at the beginning of the lecture but in the absence of an example it might not of made sense back then now it is supposed to make sense .2651

The point of doing that is that 3T would be in this case 0 here 3T will be π over 2 so very easy value to plug-in 3T will be π 3T will be 3 π over 2 and finally down here 3T will be 2 π .2662

So since I am going to be plugging in 3T , I already notice that here that means I have some very nice values here nice multiples of π over 2 to plug-in so get nice easy answers.2686

So far plug-in 3T = 0 I see I got 4cosine of 0 so 4 in the top and 5 in the bottom but im not to worry about each of the - T yet so I will just put that aside for now will figure out how that affects the solution later .2698

So you have 4 5 and the first 1 by playing π over 2 that means sin is 1 cosine of 0 so I just gets -3 and 0 here for π over 3 that means 3T is equal π so cosine is -1 & sin is 0 so -4 -5 for 3 π over 2 my cosine a 0 my sin is - 1 I get positive 3 and 0 and finally for 2 π I am back to where cosine is 1 and sin is 0 so I get 4 and five here .2717

That gives me a nice set of points to start plotting so, let me go ahead and plot those I see a start out T = 0 on it 4 5 so there is 4 and 5 is up there C = π over 6 assume it -3 0 so this is We flag this as T = 0 π over 6 is at -3 and 0 so there is π over 6 .2757

π over 3 is at -4 -5 so, there is -4 -5 there is where we are at π over 3 and π over 2 were 30 and at 2 π over 3 we come back to where we started to π over 3 come back to where we started more starting to repeat ourselves.2787

So I can see a nice ellipse forming here we see if I can connect these points up to make a nice ellipse .2820

And I also know the direction of travel because it started out T equal to 0 and it goes down Southwest here.2827

So that is heading down and spirals back up back to where it started so, that is kind of the basic shape of the graph but there is 1 element that I did not incorporate here which it well actually there is I didn't look at the other solution.2843

If you graph this solution here you just can get the same graph sort of rotated around by a little bit so , go ahead and try it if you are instead plug in some values of T and you will see that you end up getting the same ellipse that is kind of rotated around.2860

So let me now incorporate the other missing element which is this, E in - T and so E - T goes to 0 when T gets large so that makes the this ellipse Smaller and small smaller so it makes it instead of following up your ellipse an elliptical spiral getting smaller and smaller and makes it spiral in origin .2876

I will do my best to draw that for you mix a spiral in origin so instead of just following the ellipse purely getting smaller and smaller as it goes around. 4844 Still following the same basic shape of the track but it is getting smaller and smaller as it spirals in in towards the origin.2907

There be bigger versions to that or they are all kind of following this general elliptical shape but they are all getting hoops are all getting smaller and smaller and spiraling in.2937

So to draw a few more those spiraling in all going counterclockwise and getting smaller and smaller as they do it .2951

Slowly recap how we figure that out first thing I just picked 1 of the solutions you could pick E 1 I just picked the first 1 but you do not need to look at both that is a nice part and then I noticed that I had that 3T in there and I wanted to make that the multiples of π over 2 .2968

Because I know I have multiples of π over 2 fit into sin and cosine them and get very easy values to plot.2985

So, that is why I set 3T's π over to will and T would have to be π over 6 so that is why, I pick multiples of P over 6 down here is my x and then that meant that when I multiplied each 1 by 3 I got multiples of π over 2. 2992

And so when I plug-in those multiples of π over 2 π into my cosine and sin, I got these points for 5 -3 0 -4 -4 5 +3 0 and then asserts repeat itself it for 5.3007

Sets was graphing when I got that point that point this point in this point and it start repeating itself over back beginning .3023

So, I connected those up I got ellipse which is the blue graph right here now that blue graph is not the final graph that is not the final shape of the solution trajectories because, we have this other piece we have to incorporate this in the - T .3032

Everything is getting multiplied by E - T as it as it circles around here and that E- T is getting smaller and smaller so, it is kind of dragging the whole thing down the 0 which means a set of a single ellipse.3048

You got it getting this shrinking and shrinking as it goes around and around which really means you have this elliptical spiral in origin.3063

That is what I was drawing with these red curves here is getting smaller and smaller and it is shrinking down in origin.3070

Such a happy without 1 in our next example here were going to solve all the following system and were going to graph the solution trajectory.3078

This is X = X′ = 3 -5 5 -5 you will notice that that is the exact same 1 as before sort of start with her general solution from before .3092

But now we also got this initial condition , X0 = 615 so, we are going to use that initial condition to find the values of the constants in the general solution from before .3102

So, we write down what the general solution was the general solution I am just quoting this from my think was example 3 where we solve this so if you just tuned and for example five here you want to go back and look at example 3 and you'll see where we solve the same matrix.3114

We did not have the initial condition but we just use the matrix to get C1 E - T x 4 cosine of 3T -3 sin of 3T /5 cosine of 3T + C2 E - T x 3 cosine of 3T +4 sin of 3T /5 sin of 3T.3134

So that was our general solution from example 3 using the same matrix of that star general solution.3171

I do not want to rehash all that you can go back and look at the video for example 3 what were going to do is work in a plug-in T = 0 for any use that value of T = 0.3180

So everywhere through here in a plug-in T = 0 and were these the fact that E 0 is 1 cosine of 0 is 1 & sin of 0 is 0 so all the sin terms are to drop out all the cosine terms or disbelievers with 1 so exit 0 plug-in 0 everywhere is C1 x in the 0s that is just 1 for cosine that is 4-3 sin which is 0 5 cosine is 5+ C2 x 3 in the 0 again drops out since 1 3 cosine is 3+4 sin is 0 5 sin a 0.3190

Okay so, that is supposed to be equal to 615 and what that really is doing is it is encoding a system of 2 equations and 2 unknowns.3238

So, if I write out the first line there I get 4 C1 +3 C2 = six bottom line says 5C1 +0 C2 = 15 .3249

So, that is obviously a pretty easy system to solve the bottom line right there tells me that C1 = 3 if I plug that in up here I get to 12+3 C2 = six and so I would get C2 = - 2 .3264

so it gives me my C1 and C2 Upland is back in general solution .3281

Is my C1 was 3 E - T x 4 cosine of 3T -3 sin of 3T over 5 cosine of 3T.3288

Now my C2 is - 2 so, -2 in the - T x 3 cosine of 3T +4 sin of 3T over five sin of 3T and I think I can combine those together so I will have an E - T everywhere and if I look at the top I see I got 3 x 4 cosine.3305

So that is a 12 cosine over here, I've got -6 cosine so that please me and with a net of six cosine 3T now -9 sin and -8 sin over here because of the - 2outside so -9-8 is -17 sin of 3T .3338

And then in the bottom I see I got 15 cosine 3T -10 sin of 3T.3367

So that is my solution and I am in a try to graph that and remember you want to graph it where you plug in multiples of π over 2 in sining cosine .3383

So discuss a 3T is 0 3T is π over 2 3T is π 3T is 3 π over 2 and 3T is 2 π , you see what happens with each 1 when 3T is 0 the sins dropout I get the cosine 5708 I get six and 15 right that is a fraction it is really a vector they're so 615 π over 2 the sins give me 1 so -17 -10 π will give me the cosine give me-1 so -6 -15 and the 3T over that the 3 π over 2 give me the sin give me -1 so 17 and 10 .3398

And the 2 π will give me the same as if we're plug-in 0 so be six and 15 again.3470

So, let me set up some axis and try to graph that will substrate it all.3479

That looks bit better so all graph this by 3s, 3 6 9 12 15 18, 3 6 9 12 15, 3 6 9 12 15 18 , 3 6 9 12 15 18.3488

So, start out at 6 15 20 graph 615 six down six over horizontally there is my starting point there now -17 10 90 17 just short of -18 10 is just up from nine that I go to -6 -15 so and up down there .3521

And finally, -17 -10 is -17 that should have been -10 so that takes me down right there and 17 10 is your 17 and here is 10 .3548

And so I seem to have basically an elliptical shape because we know that it starts repeating after that.3576

however there is 1 aspect of this that I have not taken into account which is that there is an E - T so that E - T what happens as it takes that ellipse And it depresses it as it spirals around .3588

so each each time his browse around it is going to get smaller and smaller.3606

So let me draw the true graph and read it does start there is T = 0 there but it gets smaller and smaller each time it travels around the origin.3611

So that red graph is this true solution trajectory system .3626

So it is go back and see what we did there we started out with the solution matrix equation which is something really figured out an example 3 so, he do not know where that came from and go back and look at example 3 and you will see where we got this general solution.3642

So I just copy that over from example 3 and we plug-in T = 0 because that was the initial condition were given which met the E- T .3658

Those alternative to 1 the cosine the 3 T those alternative to 1 and so those are my thoughts right at the 4 and 5 here and that is forgot the 3 here.3666

And that all the sin term sin is 0, 0 so those all kind of dropped away when we plug-in T = 0 .3679

So I got C1 x 4 5+ C2 x 30 = 615 , and that resolves into 2 equations in C1 and C2 which are very easy to solve C1 right away plug that back in plug back in plug C1 back in first equation we get C2 so I put the C1 and general solution and the C2 in general solution.3686

And then I gathered like terms I got all my cosine's together all my sins together which is where I got the 6 in the 17 that six for example came from 3 x for which is 12-2×3 so 12-6 give me that 6.3712

That 17 came from 3 x -3 and -2×4 so that is how I got -9 -18 -17 that is where that came from .3727

Then the 15 cosine 3T and 10 sin 3T that just came from the bottom of each 1 of those vectors that was my actual solution and then I graph of plug-in values of T that would make 3T equal to multiples of π over 2 .3740

So 0 π over 2 π 3 π over 2 and 2 π of plug-in those values of T, and I got specific points here and I plotted each 1 of those points there is the first 1 there is the second 1 there is the third 1 here's the 4th 1 and then it started to repeat itself after the 4th 1 .3757

I drew my first blue ellipse Which is what the solution would look like if there were no E the - T there but since there is even - T there , I know that ellipse is getting smaller and smaller as it travels around so it is not a true ellipse it is an elliptic spiral it is getting smaller and smaller and spiralling in towards the origin .3780

So that red curve there is our actual solution for which describes the solution trajectory system .3801

So that is the end of our lecturer on complex eigenvalues and next lecture were to talk about what to do when you have repeated real eigenvalues.3809

This is all part of the series on systems of differential equations and in general this is part of our series of lectures on differential equations here on educator.com .3820

My name is Will Murray, thanks for watching and will see in the next lecture, bye bye.3831

Hi and welcome back to the differential equations lectures here on educator.com, my name is Will Murray and were working on systems of equations and we already had a couple of lectures where we studied systems of equations where you look at the eigenvalues E matrix.0000

We had 1 lecture where you talk about where we talked about 2 distinct eigenvalues and we just had lecture where we talked about complex eigenvalues 0014

In this lecture, we are going to talk about repeated eigenvalues so, what happens when the eigenvalues, the matrix and you discover that you got a multiple root of the characteristic equation.0024

That is what were going to analyze today because it is kind of behaviour from anything we seen before so, it is go ahead and jump right .0036

Just remind you where we are to solve the system of linear differential equations X′ = A X .0043

A would be a matrix here so, is using it 2 by 2 matrix the way you do that is you find the eigenvalues and the eigenvectors of the matrix A .0054

So, will talk about in this lecture is what happens when the characteristic equation has repeated root so, lowers you have a repeated eigenvalues and so, let me show you how to handle that.0062

First you find a corresponding eigenvector B so, just like we did before you try to solve A- RI x B = 0.0074

That is exactly what we did before then this parts new where to find so, method called generalized eigenvector W which means A- RI W = not 0 but V .0082

So, that is the then new element here is that instead of trying to solve A- RI time something= 0 will solve that first will get a V and then we will take that V and will plug that in on the right-hand side and try to find this W such that A- RI VW = V .0094

Remember to call that the generalized eigenvector.0121

Generalized eigenvector basically when you hit it with the A- RI matrix what you get is the original eigenvector you will get 0 get the original eigenvector .0125

So, let us see how that plays E solution reform the 2 principal solutions the first 1 is exactly the same thing we did before.0137

It is E RT x B so, that R is the eigenvalue and that V is the eigenvector just like before.0151

That is exactly the same as we are been doing it the last couple of lectures here is the new twist.0162

The second principle solution is this entire thing T E RT x V+ E the RT x W .0170

So, these the same eigenvectors before the W is the generalized eigenvector.0179

That whole thing put together is our second solution so, it gets kind of complicated and the general solution is what you get by tacking a constant onto each 1 of those and then adding them up so, I just plugged that in here.0185

That C1 x E the RT the plot + C2 x at entire thing E RTV + E RT W and then just like all the other all the other problems use initial conditions to solve for C1 and C2 if there given.0201

If you are given initial conditions you just stop your with the general solution if you give an initial conditions you use them at this point the very end to soften the C1 and C2.0218

So, there is 1 more thing we have to do with these systems which is graphing the solutions so, the way this works out let me remind you what the general solution looks like it looks like C1 E RT x the first eigenvector V + C2 x this whole quantity T E RT x V + E RT x W .0228

So, the first thing to do here is just kind of forget about the second solution just look at C1 E RT x V and when you graph that when you graph that what you get is a sort of straight line of solutions.0259

That a straight line of solutions only a line through the origin and then depending on whether R is + or - they E be expanding out a long outline or contracting E origin along that line so, they might be expanding out or this another color they could be contracting in towards the origin depending on whether that Eigen value is + or - .0281

Then the other solutions swirl around this line and it kind of depends on whether you looking at + or - values and C2 so, what is going to happen is if C2 is + than they R going to approach + multiples of V and C2 is - then approach - multiples of V .0305

Let me try to show you the 2 general pictures and then we will see some more specific examples later.0328

So, if C2 is + if R is + then the line is sort of expanding and if if this is V if this is the right here then all the all the other solutions are sort of trying to approach their swirling around and whenever C2 is + they are trying to approach + values of V.0336

Whenever C2 is - they are trying to approach - values of V so, get the sort of spiral swirly pattern there is a straight line E middle and everything else is swirling to 1 side or the other .0370

So, we will see an example of this so, that is the case when you have a + eigenvalue or rethink sort of expanding as it progresses.0389

Let me talk about the case we have a - eigenvalue you will still have that fixed axis defined by the first solution there the first eigenvector but if R is - if R is - then these things will be shrinking down to 0 .0398

And so, again when C1 when C2 is + that although be going into 0 but the sort of spiral around and try to approach that first eigenvector so, those were spiralling like this is.0419

It does not go round and round it just kind of swirls in and then when C2 is - get that away from the edge of the screen when C2 is - else swirl in to the - multiples of the eigenvector .0446

So, that is fairly subtler probably be a little easier to understand .0470

Check out some examples and do both computations and the graphs and you will see how to make these things on your own .0484

So, first example here we find the general solution to the following system X′ is 0 1 -4 and 4 .0490

Now remember all of these matrix systems of differential equations it all start out the same way got a find the eigenvalues and the eigenvectors of the matrix so, to find the eigenvalues remember start by subtracting R along the main diagonal .0499

So, will get a quadratic equation from that if I cross multiply I get - R x 4 - R - -4×1 so, that is + 4 = 0 so, see about it R ² - 4R + 4 = 0 so, R - 2 quantity ² = 0 so, I get and R I get a double root both of my roots R equal to 2 so, and this repeated eigenvalue situation .0514

So, first of all I'm going to find my eigenvector corresponding to R = 2 so, 0 0-2 is - 2 and 1 and 1 and -4 and 4-2 is 2 and I am trying to find an eigenvector that when I multiply that matrix by it I get 0 .0548

Remember and easy way to do this is to fill in 1 for Y and see what the X should be in order to make that equal to 0 and I see that it would have to be if I put a 1 half here than -2×1/2 E - 1 + 1 = 0 .0573

So, my eigenvector would be 1 half x 1, by the way if you do not like my way of writing in a 1 for the Y you can also, solve this using free parameters .0589

So, we practice that back in our review of linear algebra few lectures ago I do not really like this in this fraction in here and remembered eigenvector you can multiply by a number by scalar and it will still be in eigenvectors.0603

To get 1 and 2 for my eigenvector and that is it for my eigenvector but now I have to find this generalized eigenvector will remind you how that works you try to solve A- R I x W = V where V is the eigenvector you just found.0619

So, I am in a try to solve that now for W A- R I is the same matrix is before so, you write down what I am doing here trying to find the generalized eigenvector.0641

So, I got to the same matrix as before is it is the same A- R I -21 -42 x my W which I will does call X and Y for now X and Y = the eigenvector none 0 anymore but 1 2 .0654

So, right this out I got - 2X + Y = 1 -4 X +2 Y = 2 .0679

Those equations are redundant and you can get 1 from the other 1 and other that is a general principle here that that ought to be true if not then we would have made some mistakes some risk across out that second equation and look for any X and Y that satisfy the first equation.0692

The easiest 1 I can see there would be X and Y = 0 and 1 if Y is 1 in X0 then that will be a solution to that equation.0706

By the way I mentioned when you are finding eigenvectors which can multiply it by a scalar if you like to make the numbers nicer that is right in here multiply that V by 2 in order to make the numbers nicer that is not true with the generalized eigenvector.0720

You cannot multiply by scalar's anymore because it would no longer be a solution to this equation so, if you do not like the numbers you get for the generalized eigenvector you kind of stuck with them you have to keep going with that generalized eigenvector .0737

Now that I found the eigenvector and the generalized eigenvector let me go and plug goes E general form of the solution which was all remind you what it was in general.0752

It was C1 E RT x V + C2 x ERT x V + E RT x the generalized eigenvector W .0767

So, I figured out all the quantities you are just need to drop them in and I will get my solution so, in this case my C1 and do not know what that is just I just leave it there my R is 2 came from over here so, E 2T now my V is 1 2+ C2 x T E 2T my V still 12+ E RT E 2T and my W is 01.0786

And that is my general solution to the system .0821

So, let me show you a quick review of how that worked out we always start by finding the eigenvalues in eigenvectors of the matrix they subtract R down the main diagonal cross multiply here + - that is when doing here across multiplying - 4 give me a +4 get a quadratic equation turns out has double root which means I am in this situation repeated roots .0837

So, I take that is that the 1 word that we got and plug it back E matrix's meaning subtracted down the main diagonal lines I had a - 2 and 2 that A- R I .0863

And I try to find the eigenvector which means installing for a vector sets it when you multiply that matrix by that vector you get 0 start out by putting a Y in his 1 might figured out that X had to be 1 half to make that work .0879

You could also, do that with free parameters if you are more comfortable with that method from linear algebra so, i got this eigenvector 1 half 1 did not like the fractions I multiplied by 2 to get 1 2 .0896

Those my eigenvector and if generalized eigenvector in solving this equation A- R IW = V so, I A- R I is the same matrix there is my W on trying to find that and there is that V that comes from the eigenvector I just found .0909

So, that turns into 2 equations but 1 of them is redundant so, I got rid of the second 1 and I just try to find any X and Y that works and I got X and Y to be 0 and 1.0925

Again you could solve that with free parameters if you like but it is really such an easy 1 that I do not think it is necessary.0937

So, that I took all that information I plug it into my generic general solution for so, had this back at the beginning of the lecture in the overview got this generic general solution for so, plug-in R = my eigenvalue my eigenvector my eigenvalue again my eigenvector again it was that eigenvector.0943

That eigenvalue, eigenvalue again and then finally my generalized eigenvector of plug that in and I get my general solution .0965

So, hang on to this week is really use this exact 1 E next example were to learn how to graph this.0973

So, for the second example here were the graphs and solution trajectories to the previous system of equations so, I got written down here is the exactly the general solution to example 1 so, the escorts R looks like a big mystery then go back and watch example 1 because that is where we figured out this general solution here .0981

We start with the matrix we found the eigenvalues in eigenvectors and we eventually came up with this general solution .1005

Let me show you how the graph plays out .1014

These graphs are little bit complicated does take some practice and working through some examples before you get comfortable with him .1019

So, let me put a scale on here and it is all sort of motivated by the eigenvector and the eigenvalues that is really what you want to focus on is the eigenvector E eigenvalue .1027

What I am first going to do is just graph this first solution C112 x E 2T so, 12 x E 2t to go to vector 12 that is 1 of the horizontal direction and 2 E vertical direction and what that gives me is a fixed axis of solutions.1041

So, make my axis there and I notice that 2 is + which means that these solutions are growing there getting bigger and bigger and bigger.1065

and in the - direction that would correspond so, this would be C1 being + in the + direction the kind of going off to - infinity to getting bigger E -direction and that is really because the eigenvalue is + .1083

So, now I want to look at what happens when I consider the second set of solutions , these over here and so, if I start out with if I take for example C2 = 1 and C 1= 0 C2 is equal 1 at T = 0 I am going to be at 01 because the T = 0 will make the 1 2-term dropout.1099

That means I will start at 01 so, my 01 is here and it is going to grow from there what's can happen is as a grows the dominant term is the TE the 2T.1129

So, it is going to pull towards the 1 2 direction so, it is cannot pull towards the 1 2 direction .1147

So, as this grows it is going to pull out in the 1 2 direction number that was 1 2 right there their first eigenvector right there.1154

And that is can happen any time, C2 is greater than 0, anytime C2 is going great is greater than 0 it is been a start of the top left side of this first axis and its could end up pulling out E + direction .1164

So, even though the curses are down here are good and up swirling around and attracted back to that + direction.1181

So, you get these this swirling effect all of these repeated eigenvalue graphs have the swirling effect they always swirl E out if the eigenvalue is + or inward if the eigenvalue is - .1191

Now down here were looking at - values of C2 which means for large values of time they are going to be going to - when T is very large they are going to - values of 12.1215

So, that means there swirling down to the - part of the plane here.1236

Slowly draw some curves there swirling to the - playing here everything on the side of the major axis of the axis defined by the eigenvector is swirling down to - multiples of the eigenvector.1244

So, even though they start out sort of parallel to the major axis they swirl around and try to go to the - side of the major axis .1269

So, that is what the graph would look like let me again remind you how you would know this you would start out so, so, for so, you start out with the equation and we figured out the equation in example 1 1286

I did not do that right here but you can go back and look at example 1 if you do not remember where the general solution came from.1299

I started out by looking at this first solution 12 x E 2T and that is I got this major axis with the initial eigenvector and then as T gets bigger it grows and grows and grows and a C1 were - that is kind of the - of that vector and there it is going down into - territory so, there is-1 and 2 .1305

Now the trickier part is when you look at solutions with C2 we notice when T = 0 that term drops out and so, they just start at 01 so, that is that term right there that 01.1328

But when T gets bigger this 1 2-term starts to dominate and he gets pulled over E 1 2 direction and anything on this side of the the major axis has the same effect so, anything on the site has C2 greater than 0 and so, it is can end up getting pulled towards the + 1 2 direction.1346

Anything on the other side gets pulled down towards the -1 direction and that is why everything down below is swirling in that -12 direction .1367

So, all of these graphs they all have this sort of major axis swirling on each side swirling in opposite directions on opposite sides the only difference is if it is a + eigenvalue like we have here there swirling outwards if it is a - eigenvalue were to see that E next example there swirling inwards.1378

So, practice 1 like that and then you will really be aware of all the possibilities since try out that next example.1399

So, this example we got a find a general solution to ex-' = -41 -40 and remember all start with finding the eigenvalues in eigenvectors.1405

So, find the eigenvalues subtract the R and I will get -4 - R x - R - -4 is the same as +4 = 0 .1416

So, R ² - R ² + 4R + 4= 0 and we want to solve that of course that just factors into R +2 ² = 0 so, we get a double root at R = - 2 and so, we only have 1eigenvalue with repeated eigenvalue and so, plug that in a find the eigenvector .1432

So, that means -4 - R is - 4+2 so, - 2 1 -4 and 0 - -2 is 2 and I will find eigenvector that if you multiply that by that matrix you get 0 .1465

To do my normal trick plug in a Y = 1 here feel like that trick you can always do this using free parameters for - 2X + 1 = 0 if I make that to a 1 half there then that would work nicely goes I would get -1+1 = 0 .1485

So, my eigenvector is 1 half 1 but I do not like this fraction so, multiplied by 2 and I will get 1 2 for my eigenvector .1506

Now, here's where it starts to diverge from the previous examples of systems of linear equations difference here is because we got this repeated roots we cannot find another eigenvalue eigenvector combination.1520

Instead, we find the generalized eigenvector which is what we do that as we solve A- R I x W = V and more that V is the eigenvector that we party found the W is what were trying to solve for.1534

So, try to solve-21 -42 x my W which for now call XY and do not what is yet 12 and so, get - 2X+ Y= 1-4 X+ 2 Y = I see that those equations are just multiples of each other so, there is really no use in hanging on to the second 1.1551

I cannot so, I am looking for any X and Y that will solve the first 1 and again I can see that and easy XY that will work would be 01 if Y is 1 and X is 0 then not that would certainly work there.1582

Some use that as my W my generalized eigenvector and let me remind you cannot go multiplying generalized eigenvectors by scalars appear we took we had an eigenvector and we multiplied by a scalar to get rid of some nasty fractions you can do that with eigenvectors but not so, with generalized eigenvectors they are very specific things so, you cannot just multiply them by scalars.1602

Having found that generalized eigenvector of all set to go with my general solution for me remind you the generic form for the general solution we had this at the beginning of the lecture was C1 in the RT R the eigenvalue x V + C2 V the eigenvector C2 now gets confiscated T RT x V+ E RT x W .1630

So, W is the generalized eigenvector so, figured out all those quantities except for the C C1 each of it now my R is my R as was - 2 so, E - 2T x V is where my V is 1 2+ C2 x T RT R still - 2T x V is 1 2 + E RT's that is each of the same thing E - 2T now my W is 01 .1660

That is my complete general solution so, that is the end of that problem but before we move on to be moving on a drawing a graph of that but let me show you again out what the steps were solving that.1711

So, we start out by finding eigenvalues in eigenvectors so, that means to find the eigenvalues you subtract R off the main diagonal finding A- R I there find its determinant C Cross multiply there and that is what I am doing with this equation right here finding its determinant there is-4 but that is on the - diagonals.1737

That is why that turned into a + 4, their + 4 and that simplifies into a quadratic equation which is pretty easy to solve, but we get a double root it - 2 .1759

Go ahead of plug at - to backend find the eigenvector and since I had some fractions that I did not like there I multiplied by 2 which is okay for an eigenvector is not okay for generalized eigenvector.1772

Which is what were to find next so, got to make that via vector there it is a vector so, finding the generalized eigenvector you again look at A- R I the difference, though is that on the right-hand side he open 0 you put the original eigenvector and then you try to solve this generalized eigenvector W here .1786

So, got a couple equations for the X and Y and W Little is redundant so, we got rid of it and then we figured out a X and Y quickly that would work there and so, really use that as our generalized eigenvector. 1812

And then our general solution this is something that gave you at the very beginning of lectures on just quoting it at this point go back and looking up at the beginning of lecture.1825

That is where this comes from and then I just plug in my eigenvalue - 2 my eigenvector eigenvalue eigenvector eigenvalue and finally the generalized eigenvector so, that is where the solution comes from now you want to hang onto the solution were to be going farther with this really graphing it that is the next examples .1835

Just keep track of the solution and will use it right away in the next example so, example 4 graphing some solution trajectories to the previous system of equations so, I got here is the solution that we derive in previous example .1857

So, fairly complicated solution here but we worked it out in previous example example 3 so, having just watched example 3 and go back take a look at example 3 and that is where this complicated solution came from .1874

Now I want to try to graph it which is no small task set up some axis for myself here .1890

So, were to start with the simpler of the 2 solutions which is the C1 solution we start with the simpler of the 2 solutions , I see 12 x E - 2T so, there 1 2 right there.1911

There is 1 2 and E - 2T is going make that thing just steadily shrink down towards the origin that E - 2T just makes it gets smaller and smaller and smaller .1922

It is basically because the eigenvalue is - because that and any - eigenvalue is can have that affects only show all multiples of this 1 R just getting smaller and smaller and smaller make that and blacks with little easier to there.1936

I will go ahead and show the - side of that as well so, these R - multiples of that initial solution there is the - of 12 and then these R getting smaller and smaller there gradually drifting in towards the origin but I will always on a straight line.1952

That is our major axis and effects can be our only axis because remember we said for all of these repeated eigenvalue problems they always swirl around that major axis 1 way or another so, we just have to figure out which way this is swirling around .1976

So, the way we do that is we look at this second solution here so, let me now take C1 equal to 0 get rid of the first solution C2 = 1 and so, that means my solution is 1 2x T E - 2T +01 x E - 2T for plug-in T = 0 in that first part will drop out and I will just get 01 so, little start up here at 01.1990

But when I let T get bigger and bigger all these terms are going to go to 0 but I see that the dominant term here is going to be the 1 2 so, as to when affinity it is going to approach 0 but along a trajectory defined by 12.2024

So, it is going to kind of swirl in and approach 0 along that trajectory defined by 12 and that is kind of going to kind of be a general pattern whenever I have C2 bigger than 0 it is going to try to approach the + multiples of 12.2047

So, we have all of the all these curves swirling around and approaching the + multiples of 1 2 .2071

Now the exact opposite thing happens when C2 is less than 0 were to start on the lower side of the major axis were going to approach the - multiples of 1 2so, everything's going to swirl in an approach from the - side of 12 .2097

Easier for me to draw those backwards that look like it goes more towards the origin there we go.2114

So, everything kind of swirl and approaches from the - side of 12 so, this is the the there is always a picture like this where there is a major axis things are either expanding or contracting along the major axis and that depends on whether the eigenvalue is + or - .2126

Here, our eigenvalue was - 2 which means that things are contracting along that major axis and then outside of that major axis things are always swirling around that major axis in 1 way or another and is just all you have to do is figure out which way their swirling and also, whether their expanding or contracting as their swirling.2149

So, in this case we have a - eigenvalue which means are all contracting as their swirling so, swirling in the origin around that major axis.2170

So, just to recap the individual steps there we got the solution from example 3 this was kind of this kind came straight out of example 3 so, I did not work that out now if you were that comes from example 3 is the place to look.2181

But then we started out by graphing that first solution 12 x E- 2T so, there is1 to right there and were taking different multiples of that which is a whole line of that E - 2T being - makes everything along that line drift in towards the origin along that line.2197

That gives us fixed axis and then with C2 to graph the C2 solution wrote down what we would get if we just plug in C2 = 1 which means when T = 0 is first-term drops out and we just get 0 1s that is why were starting right here at 01.2217

But when T gets bigger the dominant term here is the 1 2 is not the 01 anymore , which is why it drifts into 0 along that 1 to direction and so, that is true anytime you are on this side of the major axis or the northwest side of the major axis and time C2 is + you can adrift in 0 along + multiples of 12 anytime you are on the lower side and means you have a - value of C2 which means you can adrift E 0 along - multiples of 1 2.2236

See every following this 1 right here so, that is that is how you work out that swirling out 1 E next 1 were going to see what happens when we form some what when we use some initial conditions and see how that affects things and how we can get some solutions with specific values of C1 and C2 .2272

So, see how that works out so, example 5 resolve the initial value problem -41 -40 and X0 = 38 .2295

So, let us see how that affects things let me remind you what our solution was to the previous problem because this is the same matrix that we had actually was in example 3 and example 3 we solve this in general we found a general solution . 2307

So, if you kind of just and for example 5 go back and look at example 3 and see that we found a general solution X general was C1 x E - 2T x 12+ C2 x the quantity T E - 2T x 12+ E - 2T x 01 .2328

So, that was which we did some work in example 3 to get to that point so, like a repeat that now but if you do not remember that just go back and check example 3 see it all worked out.2366

Now let us see how this works with our initial condition because this is the new part of this problem X0 = 3 8 so, i am going to plug-in T = 0 here so, get C1 E 0 is just that 1 2 + C2 x now 0 E 0×12+ E 0 x 01 and so, a lot of these terms drop out.2375

There E 0s are all to me once I get C1 x 12+ C2 now this this term completely drops out because multiplied by 0 and then I just have C2 x 01 because E 0 is still 1 and that is supposed to be equal to 3H so, I can expand this out into 2 equations if I distribute the CE vectors.2410

Looks like I got C1 +0 C2 = 3 and 2 C1 + C2 +1 C2 = 8 right away I can figure out the C1 = 3 and if I plug-in 3 there got to 6+ C2 = 8 so, C2 would have to be 2 .2435

So, I got my 3 and my 2there so, to plug those in for C1 and C2 so, I have got to see C1 was 3 E - 2T x 12+ C2 is 2 x T E- 2T x 12+ E - 2T in this copying the general solution now 01 filled in C1 and C2 .2461

I see that I could simplify a little bit because this to this 2E- 2T to combine it with this 3 E- 2T that the T′ might have to write separately so, this is what we expand out maybe this right-hand set of terms .2496

This is 2T E- 2T x 12+2 E - 2T x 12 x 01 and then I can combine my light term E - 2T gets E - 2T x 3×13×1 and 2×0 so, 3 there .2515

Then on the bottom I got 3×2 and 2×1 6+2 is 8 + T E- 2T if I multiply that this 2in with this vector I get 24 there .2544

And that is a simple as it is going to get.2560

By the way notice that it is very easy to check now if I plug-in T = 0 the right-hand side would all go to 0 and the left-hand term would give me 38 so, that is what it was supposed to be so, I can check that I got it right that is very easy.2566

So, recap what we did there start out with a general solution that I worked out in example 3 so, that general solution was a complete district you just go back and watch example 3 see it .2588

Then I plug-in T = 0 here they so, my initial conditions on plug-in T = 0 so, were sought T through E 0 so, all my E 2Ts become 1 and the T E 2T actually drops out become 0 because of that 0 .2601

So, I just end up with C1 x 1 2+ C2 x 01 and that resolve itself into two equations, in two unknowns are very easy equations I get my solutionsC1 = 3 C2 =2 so, plot those back in here C1 is 3 C2 is 2 plot those back in here and it is a little bit messy so, I want to maybe expand out this 2.2619

Distribute it to here and here and I see of got E 2T term here and used 2T term here so, I combined those had to distribute the 3 into the 1 and 2's that turn into 3 and 636+210 010236+02 gave me 38 .2648

Then, this to distributing E 1 and 2 gave me that 2 and 4 that is my final form of the solution there but it is also, easy to check if I plug-in T = 0 in here this term would drop out because of the T outside that T outside this may turn into 1 .2674

I get 3 8 so, it does check that I match the initial condition that I was given.2689

Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues.2694

This is all part of a larger lecture series on differential equations here on educator.com .2708

My name is Will Murray and I thank you very much for watching, bye bye.2713

Hi and welcome back to the differential equations lecture here on educator.com .0000

My name is Will Murray and today we are going to talk about inhomogeneous systems and we are going to study 2 methods of solution for inhomogeneous systems.0005

Today we are going to talk about undetermined coefficients there is another method called variation parameters, we are going to talk about in the next lecture So, if you are looking for a determined coefficients you are in the right place.0015

If you are looking for solving things by variation parameters then, just wait to the next lecture because that is a totally different method of solution.0026

So, let us take a look at what we’re talking about here.0032

I will also, solve the inhomogeneous system of differential equations vector X′ is = a x vector X + GMT.0036

So, the idea here is this this a would be a matrix So, usually be a user 2 by 2 matrix of Constance vector X is the functions that we are solving for that is that is a column vector X1 and X2 .0047

Then this vector Dt would be some extra functions that we did not have before this what makes it inhomogeneous this GMT So, that is that is really what inhomogeneous means is that we have that GMT .0062

So, some extra functions will be added on the outside there and today we are to learn how to solve something with of that form .0077

So, the idea here is that first you want to solve the corresponding homogeneous system X′ = a X and you can do that by the methods of the previous lectures So, you can get a homogeneous solution C1 x vector X1 and C2 x vector X 2 .0085

So, it had several lectures now on how you solve that member that was the whole business of finding the eigenvalues and the eigenvectors of the matrix A and then plugging them back in and you get C1 E RT x the first eigenvector C2 in the second RT x the second eigenvector so, we had several lectures on that.0103

What you do when you have real distinct eigenvalues what you do when you have complex eigenvalues and what you do when you have real repeated eigenvalues.0125

So, if you do not remember that, maybe check back in those lectures because we’re going to be using that material sort of as a first step today to solving these inhomogeneous systems, the very first step is to solve the corresponding homogeneous system just as we did before.0134

So, see what happens after that for the inhomogeneous system we are to look for a single is called a particular solution which call X R to the inhomogeneous system and the idea is you look at this GMT look at the extra functions that are given in your equation and you guess something of the same form but with generic coefficients.0151

So, for example if GMT we are say 3 E to the 5T and 2 E to the 5T will be of you would be a vector there E to the 5T So, GMT is always a vector because these are these are systems of equations So, he would think of that is being vector 32 x E 5T.0171

So, that when the GMT that you we are given you would guess something that is just like that except you would be specific about the 3 2 so, you would guess you would guess something of the form of multiple of E 5T except that you would guess generic coefficients A and B instead of 3 2.0198

So, then what you would do and that is you plug it back in and try and solve for the A and B, will see some examples of it and you will get the hang of it once you figure out what your particular solution as once you plug it back in and figured out what those A and B are then you are going to take that particular solution and plug it right back in here, which means you can add it on to the general solution and you are going to get I should not have labelled this homogeneous session of labelled this the general solution So, fix that right there.0220

Next the inhomogeneous solution So, the general inhomogeneous solution is C1X1 + C-2 X2 + the particular solution .0259

The realized that new part of this lecture is really making that guess right here and solving for those undetermined coefficients and that really will make much more sense when we do some examples .0275

So, separated to some examples for the following inhomogeneous system for similar to do is just solve the corresponding homogeneous system so, what that means is the when resolve the corresponding homogeneous system and means you ignore the inhomogeneous part of the equation .0288

You ignore this part will just ignore that and will solve the corresponding homogeneous system which means really use the techniques we learn several lectures ago so, if you do not remember how to solve this, go back and check a couple lectures ago when we first started solving systems of differential equations.0304

The way you do it is you find the eigenvalues and eigenvectors of the matrix So, what we will going to do 1 - R and 1 - R doing A- R I your subtracting RI down the main diagonal of the matrix .0322

So, now I got 1 - R ² 1 - R x 1 - R -4×1 is = 0 and that is quadratic equation to get R ² -2 R +1-4-3 is = 0 this with factors nicely it is R -3 x R +1 is = 0 So, R is that 3 or -1 those my 2 eigenvalues .0336

For each 1 of you find an eigenvector by plugging that value of R back into A- R I So, R is = 3, plug that back in 1 - R is - 2 41 -2 and look for an eigenvector there so, my eigenvector I plug and Y = 1 and try to make that = 0 that is my little shortcut way of finding eigenvectors .0375

If you do not like it then there is a way to solve this using free parameters as a way learned your algebra signal and use that if you like it better, but I see all gets I want to see what the first component should be to make it = 0 and I see if I put the 2 there then when I multiply that matrix by this eigenvector I will get 0 .0403

So, that is my eigenvector corresponding to R = 3 for R = - 1 that means I am adding 1 of down the main diagonal so, get 2 4 and 1 and 2 get them to put a - 41 down the Y component and I see that I would have to put a - 2 in the X component to make that come out to be 0.0423

So, now got my eigenvalues and my eigenvectors and I just drop those into my generic formula for the general solution for the general homogeneous solution it is C1 x E RT in the R1T x the 1+ C-2 x ER 2T x V2.0446

We saw this a few lectures ago and so, this is C1 well , my first eigenvalue -1 is the first 1 So, E - T x -21 and its + C2 x E 3T x 21 and that is my homogeneous solution .0471

We use that to add on to the inhomogeneous part in the next example but let us recap what happened with this example .0497

It says solve the corresponding homogeneous system which means we are going to ignore the inhomogeneous part for now what will taken it in taken into account in the next example So, working to find the eigenvalues and eigenvectors of this matrix .0509

Subtract R down the main diagonal and then take the determinant So, cross multiply get a quadratic equation there which factors easily get 2 eigenvalues for each 1 we plug it back in the matrix and solve A- R I x the eigenvector is = 0 So, my shorthand way of doing that is to fill in 1 for the Y component and figure out what the X component should be .0525

I get my 2 eigenvectors and I just drop them in of the eigenvalues as exponents of E and the eigenvectors in as coefficients and that is my general homogeneous solution .0550

I saw to solve the inhomogeneous system that is what we are going to do in the next example .0565

So, the next example we are going to solve the inhomogeneous system which means we are going to guess something of the form that is comparable to the inhomogeneous part of the equation .0569

So, that that right there we are going to guess something that looks like that but has generic coefficients sort in a guess X is = AB x Et .0585

Plug that back into the system but will need to know what ex-' is the derivative Et is just Et was just going to be A x AB x E T so, plug that back in plug into those into the differential equation So, I get AB x E T is = 1411 x AB E T + -12 -12 x ET T .0600

I see right away every things got E T So, cancel those out right away and get AB is = now 14 xAB is = a + 4B and A+ B that is 1 x AB + -1 and 2 is going to turn into a pretty easy system of equations right away.0642

A is = a + 4B - 1 B is = A+ B +2 .0666

So, this is a system of equations 2 equations 2unknowns So, you can really solve this using any method you like from high school algebra for this 1 turned out to be really easy we just get the A's cancel there So, 4B -1 is = 0 that tells me that 4B is 1 So, be is 1 4th .0677

Then, on the second 1, that these cancel and get A+ 2 is = 0 So, a is = - 2 and so, my particular solution I guess AB even now I know that a is - 2 B is 1 4th and I have E T there and Ss, my general solution is the homogeneous part X homogeneous + the particular part that I just figured out.0699

Now the homogeneous part I figure this out on the previous example so, if you did not just watch the previous example go back and look at example 1 and we will see where this comes from.0737

You will see that for example 1 what we got was C1 x E - T x the vector -21+ C2 x E 3T x 21 now + the particular solution.0749

Now, this does not get multiplied by a constant this is very important particular solutions to inhomogeneous equations you can just multiply them by arbitrary constant So, you know I am not So, thrilled about that 1 4th right there really like to maybe multiply by something in clear denominators I cannot do that because that would make it not be a solution anymore.0777

So, I am stuck with that 4th -2 1/4 x E T So, really emphasize there is no arbitrary constant there those - that -2 1 4th are absolutely fixed, I cannot change them and I cannot multiply by anything to clear that denominator .0797

But having done that I am now finished with that example I have got my complete solution to the homogeneous system So, recap how that worked out.0825

For solving the inhomogeneous system really the first thing to do is to solve the corresponding homogeneous system that is what we did back in example 1 so, really the first Here was to find a homogeneous system .0842

To find the C1 E - T + C-2 E 3T x 21 you will see the derivations for that back in example 1 what we are doing here in example 2 was finding the particular solution to the inhomogeneous equation.0853

Which meant we guessed which I saw over here and E to T So, I mimicked it here for generic coefficients undetermined coefficients a and B and then when I take the derivative E T is just itself I get the same thing .0870

When I plug that in X′ plug of this and for X, and I try to simplify my equation right away the E T's go way I multiply this matrix x this vector which is how I get this column vector and then I separate that out into a system of linear equations.0886

Now , you can solve the system of linear equations using several different methods that you learned in high school algebra this 1 solve really easily because we had several variables cancelling right away.0908

So, got a value for A and B and then I plugged those back into my particular solution that was coming from the original guess but now I have determine what my coefficients would be .0921

That is my particular solution , I cannot multiply that by arbitrary constant am forced to just take it and I plug it back in here not there but over here is where the particular solution got plug-in add that onto my homogeneous solution and I get my complete answer there.0931

So, let us keep going with another example here we got the following inhomogeneous system and we are being asked to solve the corresponding homogeneous system with that really means is we are not can worry about the inhomogeneous part for now .0952

The next couple of examples we are to come back and we are in a solve the inhomogeneous system So, we will see that in examples 4 and 5 but we do not want me to worry about that right now we just need solve the corresponding homogeneous system .0967

Remember those all work the same way your finding eigenvalues and eigenvectors subtracts and find A- R I will take the determinant there So, I get 5 - R x 5 - R - -3 x -3 that is -9 is = 0 .0987

And that is R ² -10 R now +25-9 is +16 = 0 factors nicely R = to only show the factorization R -2 x R -8 is = 0 So, R is = 2 or 8 .1009

So, for those of the 2 eigenvalues each 1 of those we are in a plugged back in for R and get it corresponding eigenvector So, in R = 2 I am doing 5 - R main diagonal assuming it 3 -3 -3 5-2 is 3 and 1 multiply that by some eigenvector and get 0 .1038

Do my usual trickle plug in a 1 for the Y you do not like that you can use free parameters to solve the system but plenty and 1 for the Y works very well and I can see that I would have to have A1 for the X to make that work 1 for the X to make that work.1062

So, got my eigenvector corresponding to R = 2 R = 8 then I subtract R down the main diagonal there which is getting to me -3 -3 - 3s everywhere 5-8 is -3 and the put a 1 in for the Y .1080

I see that I would have to to make that work out since I got done - 3 x something -3 is = 0 that for something to be -1/2 in my eigenvalues and eigenvectors and I am in a drop those into my generic form to get my homogeneous solution .1103

Can be C1 x E of the first eigenvalue x it corresponding eigenvector 11+ C2 x E of the second eigenvalue x its corresponding eigenvector -11 .1122

That does it for the homogeneous solution remember when I can worry about the inhomogeneous part of this equation until the next couple of examples So, what I did there was I kind of ignored the inhomogeneous part.1138

I just focused on the homogeneous parts which means I need to solve for the eigenvalues and eigenvectors major subtracting R down the main diagonal and then cross multiplying to get the determinant so, that is how I got this quadratic equation here.1157

Which factored really easily and I got at R= 2 or 8, and then for each 1 of those a plugged in and I found A- R I .1173

So, plugged the 2 back in 5-2 is 3 that is we are that 3 and that 3 came from then I try to guess a vector that would make it come out to be 0s I plugged in Y= 1 and I figured out that X had to be = 1 in order to make that come out to 0 sending articles 8 5-8 is -3 So, that is for those come from .1183

And again, I plugged in my = 1 and then I realized that if I put at X = - 1 that would give me 0s and confirm that that is an eigenvector that I just dropped my eigenvalues and my eigenvectors into the generic form for the solution and I get my homogeneous solution .1205

I hang onto that will be using that later but in the meantime in the next examples we are to try to solve for the particular solutions to the inhomogeneous problem So, stick around for that.1224

So, example 4 , you find a particular solution to the inhomogeneous system is the same 1 we had in example 3 So, you might go back and check example 3 see how we solve the homogeneous problem.1237

But now, resolve the inhomogeneous system but I see when I look at this I got some terms with E to T and I have also, got some terms with E 3T and he should think of those beings we are fundamentally different .1251

So, what we are going to do is in this example we are just going to try to solve for the E T terms So, let me separate this into an E 3T term and and E T term .1263

we are just going to worry about the E T terms as he was in the 3 T OF 0 and 5 and 4 in the T I would have 2 and -5 E T .1279

So, you should think of those beings we are fundamentally different functions there and So, what you do is to solve for particular solution corresponding to each 1 separately.1292

So, in this example we are just going to solve the ET terms and then example 5 that is what we are going to handle the E 3T terms .1304

That is going to be example 5, and what we will do right now is to solve for the ET terms so, see how that works out we are going to guess an X = something of the same forms on right AB E T but I do not know what those A and D are yet I leave those coefficients generic .1314

That is why it is called undetermined coefficients to find my X′ that is the derivative of that but ET its derivative is just self So, A.B. 1335

I am in a plugged each 1 of those into the differential equation X′ and X So, I get A B this is the X′ of writing down right now is = the matrix 5 -3 -35 x my X is A.B. E T + remember of just trying to solve for the terms we are E T so , put 2 -5 will count will account for the other terms later E T .1347

I see now that it really I have has ET cancels off if you do that by multiplying by E - T if you liked and now I see that that matrix I can write it as 5 A- 3B and -3 A+ 5 B .1381

So, now I can sort this whole thing out into a system of equations on the top I got a is = 5 a -3 B+ 2 and B is = - 3A + 5B -5 .1403

So, let me try to simplify that a little bit if I move that a over to that side and move the 2 over to the other side I get 4 A - 3B is = now -2 the multiplex I moved it over to the other side if I bring that be over and move that -5 back I will get - 3A + 4B is = 5 .1422

So, I got this system of equations and indexing not a very nice 1 in incidents go work out my 6th practice it before but it is it is a little ugly kind of along the way but basically it is a system of 2 equations into unknowns So, you can use whatever technique you learn in high school algebra to solid there is lots of different techniques.1463

Leave substitution but there is other techniques you might use that would debate basically they all work to solve this.1487

So, what I am going to do , is take that first equation and I am going to say 4A is = move the 3B over the other side 3 B- 2 and so, a is = 3 4ths B / 4-1/2 and then I am going to substitute that into the second equations to get a little messier.1495

Warning you in advance, it could be a little ugly So, -3×3/4 B -1/2 of substituting in for a + 4B is = 5 and So, that is - now -9/4 B +3/2+ 4B is = 5 now 4B- 9/4 that is 16/4-9/4 that is 7/4 B and if I move that 3/2 over the other side 5-3/2 is 10 half -3/2 7/2 and that actually is a very nicely because it turns out that B is = 2.1518

So, kind of ugly in the meantime, in the intermediate steps but it worked out nicely and then remember we have 4A is = 3 B- 2 So, I use that So, 4A is = 3 B- 2 3 x 2-2 which is 6-2 is 4 So, 4A is 4 and so, A is = 1 and So, my X let me remind you was A.B. x E 3T now third out what A and D areA is 1 and B is 2 .1574

We still have not E T there So, that is my particular solution on the subscript on there to remind you that that is the particular solution corresponding which would if we plugged it into the equation would give us the E T part of the inhomogeneous terms there.1615

So, it is it is 12 x E T. Let me emphasize here there is no extra constants here there is no C1 or C2 So, no more constants that 1 and 2 are very much fixed you are not allowed to multiply this thing by any arbitrary constants because it would ruin the fact that it is a solution there.1637

So, that is the end of that example we are going to revisit this later and solve that last term the 05 E 3T.1660

Let me remind you what we did here we guessed we we are trying to solve for this term the 2 -5 E T So, we guess something that looked just like it A.B. in the 3T except we left the coefficients generic the A.B. and set up to a -5 .1668

We took its derivative which because it was just E T did not change anything we plugged it back in there is the X′ there is the vector X and I plugged it back into the ' there the plugged it back in the equation just including the E T partners.1683

we are solve in E 3T part later on and then I took this matrix multiplied by A.B. and expanded that out into a vector and this gave me 2 equations and 2 unknowns, A.B. is = these equations. 2840 Now, that was a little bit of an ugly system of equations there quite aware to solve those but was nothing that you have not done in high school algebra so whatever methods used in high school algebra you can use those to reduce and soften the A.B.1702

I use substitution in the first equation of the first equation I saw for A and terms of the third is right there plugged it back in to the second equation .1734

Then I we are that go came over here and salt for B and I finally plugged that back into the first equation and solve for A.1745

So, little messy there but it be actual values of A.B. came out nicely and I drop them back into my initial guess and I got my particular solution which remember you do not want to multiply by any arbitrary constants.1755

So, the next example what we are going to do is where the try to account for this other turn this in the 3T term and R finally in example 6 will put them all together and get our complete solution to this system.1767

So, let us go ahead and look example 5 here also, example 5 this is the same system that we have been working on for the past couple of examples the difference is that we are now going to try to account for the E 3 term in E 3T term on the right-hand side .1781

Let me remind you that this E 3T term that this right-hand side this inhomogeneous part where you write this as 05 x E 3T +2 -5 x E T .1799

Separate out the ET stuff from the E the 3T stuff anything of those is completely different problems in fact E T we solved that in example 4 single back and check that out if you do not member how we did that.1818

What we are going to do right now is we are going to try to solve for this term the E 3T and so, the way we do that is we make a guess for a particular solution our X is going to be something like that So, we have E 3T .1833

Really generic coefficients A.B. and we want to plug that back in know the derivative X′ now E3T its derivative begin extra 3 there So, it is 3 x A.B. x E 3T I just got an extra 3 there from taking the derivative of E 3T.1849

Plugged each 1 of those back into X′ and X and remember also, going to include that inhomogeneous term of E 3T So, get 3 x A.B. E 3T is = our matrix 5 -3 -35 x our X is A.B. E 3T we are to try to solve for A.B. now + our inhomogeneous term 05 E 3T .1873

Do not worry about the rest of the inhomogeneous terms where he found a particular solution back in example 4 corresponding to that So, I see now in my equation got a E to E 3T everywhere.1909

So, many cancel that out in the 3T there there there the way to do that if you wanted to be really formal about it is your multiplying both sides by E - 3T.1921

Cancels that out and now I see that this matrix will turn into if I multiplied by the vector AB I get 58 - 3B and -3 A+ 5B and So, I see that I got a system of equations here write that out.1930

3A is = 5 A- 3B +0 there So, going right +0 and 3B is = - 3A + 5B +5 at 5 is coming from that term that inhomogeneous term over at the end there .1952

So, I have again 2 equations and 2 unknowns is a matter of high school algebra to try and solve these things using really whatever method you like also, simplify them a bit first .1976

If I subtract that 3A from the 5A, I get 2 A- 3B is = 0 and if I subtract that 3B over there and maybe plugged move that -5 over to the other side - 3A + 2B is = -5 So, I think I am many you substitution.1989

So, right over here , I see that 2A is = 3B So, a is = 3/2 B gets a little bit messy but I think I plan this 1 So, that the numbers will work out nicely in the end .2019

We play that back in the second equation I get -3×3/2 B + 2B is -5 and So, -9/2 B + 2B is -5 now 2B is 4/2 B So, for has -5/2 is -5/2 B is = -5 So, I see the B is = - 2 no positive 2 to make that.2036

Work that I plugged that back into a = 3 has been get a is = 3/2 x 2 which is just 3 and So, I figured out my in my B and now I just have to drop them back into my initial guess my particular solution right there was A.B. 3 in the 3T So, my is 3 my B is to x E3T.2073

Kind of looking at this guess over here and that is my particular solution corresponding to that, that if you plug it and would produce this part of the inhomogeneous to the inhomogeneous term there would produce the 05 it would not solve the 2 -5 problem but that member we solved in example 4.2102

So, let me recap what we are doing here we got this complicated inhomogeneous differential equation we solve the homogeneous part back in example I think was 3 and in example 4 we solved the ET part .2126

Now in example 5 what we are trying to do is to solve E 3T part and So, the way we do that as we make this generic guess E 3T to mimic what we are trying to find but we leave under the coefficients of determine their the A and B .2142

We had to take its derivative sorting the derivative X′ the 3 props out and get an extra factor of 3 out there and we plug it in for X′ and for X and So, for X′ got 3 A.B. E 3T for X we got to A.B. E 3T .2159

We noticeably got in the 3T everywhere So, those all go away those cancel away but this matrix x this vector expands out into 5 A- 3B -3 A+ 5B and So, on the left-hand side we got 3 A3 B on the right-hand side we got those terms + the 05 those are coming in here and here .2180

It turns into to equations into unknowns not too bad you can use whatever you learned in high school algebra to solve these substitution linear combinations determines any of those work.2202

I use substitution so, I solve for A in terms of B here it is right there that was in the first equation than I plugged it back into the second equation for a .2215

That gave me a nice equation for B I got B = 2 plug that back in here and got a = 3 that took the whole thing and plug them back into my initial guess for the particular solution 3 and 2 x E 3T .2223

So, my particular solution corresponding for E 3T terms on the right-hand side So, finally an example 6 over to do is to put all these different results back together and get a complete solution to that inhomogeneous system of differential equations.2242

So, we will see how that plays out example 6 is the same system we been studying and examples 3 4 and 5 but we want to put all those answers back together and find out what the complete solution will be.2260

So, redo all the work of the answers that we figured out in her previous examples So, an example example 3 we solve the homogeneous system which means we completely ignored all the inhomogeneous terms always in the T terms .2278

We got X homogeneous was C1 x 11 x E 2T + C2 x -11 x E AT.2294

That was in example 3 go back and watch example 3 that came from an example 4 we focused on the E T terms of right-hand side .2312

We looked at that and that and we figured out a particular solution that produced those 2 terms in that particular solution was 12 Et that was using undetermined coefficients.2327

Very important here there is no C here there is no C1 C2 C3 anything like that So, there is no constants there is no C because that is a particular solution if you change the constants you are going to be changing the answer and it will work out.2342

In example 5, that was where we found a particular solution to account for the E 3T term and that turned out to be it was 3 2 E3T and again there is no constants involved there you do not multiply that by an arbitrary constant you just leave that. 2359

We very carefully found those coefficients 3 and 2 and you cannot multiply them by anything without messing up your solution and so, what we want to do is just add all the solutions together and will get the general solution to the problem.2389

So, all I am going to do is just add up those 3 parts of the solution I still have the C1 and C2 for it from the homogeneous part E 2T + C2 -11 E 8T and now minute add-on each 1 of those X particular.2405

So, configure those of being the first particular solution in the second particular solution 12 E T and for the second 1 32 E 3T so, to the 3T So, we add all those together.2432

That is the complete solution to that inhomogeneous system.2460

So, let us recap everything I did in this example which actually was not much work it off because we did all the work in the previous example So, if you have not watched this previous examples that is where I am getting all these individual solutions for .2467

We did all the work back there in this example we are just kind of assembling them together so, we first looked at the homogeneous solution and then back in example 3 we found that the homogeneous solution had this form right here .2485

An example 4 , we looked at this part of the ET terms of the inhomogeneous solution we found a particular solution using undetermined coefficients that will produce that part of the inhomogeneous of the inhomogeneous terms .2501

Finally an example 5 , we looked at E 3T and begin using undetermined coefficients we found a particular solution that would produce that part of the inhomogeneous terms .2518

We do not multiply E 1 of those by C1 or C2 we do have a C1 and C2 inhomogeneous solution essentially that is because the homogeneous solution your setting something = 0 .2532

So, if you multiply it by constants it will still work out to be = 0 but with the inhomogeneous solution you got some terms that are none 0 and So, you very carefully arrange your solutions and if you multiply them by something that we multiplying a none number by something and So, would not be come out to work after you did that.2544

So, after we do all that all this was in the previous examples we can go back and check those out to see where those numbers came from.2563

All we are doing in this example is just adding those together just assembling the homogeneous solution the first particular solution and the second particular solution into 1 big solution to the inhomogeneous system.2573

So, that kind of wraps up our lecture on the inhomogeneous systems using undetermined coefficients, got another method called variation parameters which is a totally different method for solving inhomogeneous systems.2586

So, hope you will stick around and watch that lecture it is the very next lecture we will seem to learn about variation parameters.2600

In the meantime you been watching the differential equations lecture series here on educator.com .2607

My name is Will Murray and I really appreciate your watching, thanks very much, bye bye.2612

Hi and welcome back to the differential equations lecture here on educator.com.0000

My name is Will Murray and today we are still working on systems of differential equations .0004

Were to be studying inhomogeneous systems and said there is really 2 methods to solve inhomogeneous systems that were going to be covering.0009

One is undetermined coefficients and we talked about that E previous lecture so if you are looking for undetermined coefficients just go ahead and get back to the previous lecture .0017

We got a whole lecture on that today we are going to be talking about variation parameters which is a totally different way to solve inhomogeneous systems.0025

So, see how that works out once all the inhomogeneous system X′ = a X + GMT now remember X -prime and X those are short for a column vectors.0032

So, 2 different functions X 1 of T and X2 of T of course when there was X′ E derivatives of both of those and A will be a matrix.0049

Were just cannot do this for 2 by 2 could also do all the same things for 3 by 3s, but that just makes it much more complicated so we will just go over to 2 by 2. 0060

Then, what makes it inhomogeneous is this extra term GMT some of that is also a column of functions a column vector functions of the 2 functions in there.0069

And the fact that there is that term GMT it is not just 0 is what makes the system inhomogeneous so the inhomogeneous part refers to the fact that there is an GMT there .0081

If there were no GMT there, it would be a homogeneous system we already had several lectures on how to solve homogeneous systems.0094

Remember , that is where you find the eigenvalues and eigenvectors of a matrix and he do different things depending on whether there is real repeated eigenvalues or complex eigenvalues are real distinct eigenvalues.0101

Forgot lectures on all of those if you do not number how to do those you want to check back on those lectures it is earlier on the same lecture series .0112

The reason you want to be make sure you are up on that is because the first step to solving the inhomogeneous system is to find the corresponding homogeneous system .0119

So that is what this here is all about the first step to solving this inhomogeneous system is to find the solution to the corresponding homogeneous system .0130

What that really means is that, you kind of throw away the GMT see just look at X′ = A X and you want to solve that using the methods of the previous lectures. 0142

that is all the business about the real repeated roots E complex roots E real distinct routes where you find the eigenvalues and the eigenvectors and so on so, you go through all that business .0155

You find these 2 solutions to the homogeneous system so, you get were caught X ¹ and X ² those the two solutions to the homogeneous system .0166

Then were going to look for single particular solution which will call X par for particular to the inhomogeneous system and let me show you how that works out.0178

By the way all this so, far is Exactly the same as what we had for undetermined coefficients the difference with variation parameters is how you get that particular solution.0189

So, they both start out with solving the homogeneous system and then both of them variation parameters and undetermined coefficients they both try to find a particular solution.0199

The difference is how you find that particular solution so, let us see how you do it using variation parameters which we are going to guess is you start with that those 2 homogeneous solutions.0210

The X1 and X2 and then you can multiply them called by coefficients and instead just multiplying them by constant C1 and C2 were to guess 2 functions U1 and U 2.0221

Now we do not know what you want and you 2 are yet so, you wanting U2 are those are both functions of T functions of T and we are going to figure them out and teach you how to find those.0233

So, for the time being I am just can say to be determined so, let me show you how you can find that U1 and U2 of T because once we find the 1 U2 see you will be done of the problem.0251

So, what you do is you write the homogeneous solutions into a fundamental matrix say make the 2 homogeneous solutions column vectors and what were calling a fundamental matrix .0263

This matrix we usually denote by the capital Greek letter size that is pronounced sigh capital Greek letter size of C I am going to pronounce it sigh .0275

Then what you do is you find the inverse of size so, you take this matrix and you invert it so, you find sigh in personal video little trick to invert to buy 2 matrices E second so, do not worry about that for the time being.0288

Just know that you have to invert that matrix and then you multiply it by GMT and remember GMT came from the differential equation so, that is already given to you in the differential equation.0303

You will have a differential equation X′ = AX + GMT so, that GMT just illustrate how the differential equation drop it right here and so, you multiply sigh inverse x GMT and what you get is not the U1 U2 directly .0316

You get U1′ and U2′ you get a column vector and the 2 entries are the derivatives of the U .0338

So, what you do with that is you integrate that U1′ to get U1 and you integrate U2′ to get U2 and then just drop it back into that form for the particular solution to drop that you wanting U2 of T back into the form.0346

And we will see how that works out often there is a lot of multiplication and a lot of cancellation that goes on so, this is a kind of a long process .0362

But I think it will make more sense after we work through it with a few examples.0371

So, you take this U1 U2 and plug them back into that form for the particular solution and then you are done.0376

There is quite a bit multiplication involved but it sort of all works out so, you see it is a little bit tedious but will try it out.0385

There is 1 more thing that I want to show you which is the general solution and also, how to invert matrices so, the way to get the general solution once you got that particular solution is you just add on the homogeneous solution which was the 2 solutions that we talked about before on them.0392

And you just add that to the particular solution so, I get the general solution .0413

Then 1 more thing that I wanted to mention was that how you find the inverse of a matrix remember were to have this matrix sigh which is the 2 homogeneous solutions and we have to find the inverse of it sort of halfway through solving things by variation parameters.0417

So, here is a trick for doing that if you have a 2 x 2 matrix ABCD the way you can find the inverse is first one you find its determinant which is this AD- BC thing so, that is cross multiplying + on the main diagonal and - - on the back diagonal .0434

So, then you switch the entries around the switch A and D and you put - sins on B and C so, you get - B - C and A in any just divide by the determinant.0454

So, that is a safe way to take the inverse of 2 x 2 matrices does not work for 3 by 3s but it is really nice for 2 by 2 .0468

And you can do that when the entries or numbers are even when their functions and were going to need to do it when the entries are functions.0475

What were doing variation parameters so, keep that formula in mind and will be using that were solving things by variation parameters.0483

Go ahead and see how works out in so, me examples so, the first example here were going to were given a inhomogeneous system and want to find the corresponding homogeneous system and find the fundamental matrix .0491

Remember homogeneous system means you kind of clip off that GMT on the right and that is the GMT part right there that is the GMT .0505

So, when I can worry about that were actually come back in example 2 and saw the inhomogeneous system .0515

For example 1 were just going to solve the corresponding homogeneous system so, what we had here to write it matrix form is X′ = if I write it matrix for my C 114 -2 x X and then remember were clipping off the GMT because were just going to study the homogeneous system .0521

For now come back in example 2 and solve the inhomogeneous system and so, remember the way to solve these homogeneous systems is to find the eigenvalues and the eigenvectors of the coefficient matrix.0546

So, that is this matrix right here and so, A- RI = 1 - R 1 4 and -2 - R just subtracting Rs down the main diagonal and so, I get 1 - R its determinant 1 - R x -2 - R -4 = 0 .0560

If it looks like I got R ² + 2R - R so, that is R ² + R and then -2 and then there is also, a - 4 so, -6 total = 0 .0584

It is nice that factors really easily because that factors into an R+3 x R -2 = 0 as I get my 2 eigenvalues R = 2 and R = -3 so, as my to eigenvalues and now for each 1 have to find an eigenvector so, let me apply in R = to first R + 2 plug that into the A- RI and I get -114 and -2 - 2 is -4.0600

And find the eigenvector R looking for a vector that when I multiply that matrix by it I get 0 0 vector so, mild trick is to plug in 1 for the Y value in and figure out what the X value should be.0634

And I see here that if I make the X value equal to 1 as well then I work that means that the matrix x that vector will be equal to 0 .0649

So, that is my eigenvector corresponding to the eigenvalue R=2 you can also, find these eigenvectors were rigorously using the old linear algebra technique of finding free parameters .0657

You can do that that way if you do not like my sort of off-the-cuff way of just plugging in 1 for the Y value.0673

Let us look at the other eigenvalue which is R= -3 and if we plug-in A- RI that means I am adding 3 along the main diagonal so, get 4 and 1 and 4 -2 +3 is positive 1 .0679

And again I am in a plug-in 1 for the Y value and try to figure out what the X should be to make it equal to 0 4 X + 1 = 0 so, my X would be look like -1/4 so, I plug that in up here -1/4.0699

And I do not really like the fractions in there remember with eigenvectors you can multiply by scalar if you like to make the fractions little easier and so, I think I am going to multiply by - 4.0718

So, my eigenvector corresponding to -3 will be multiply by positive for to get rid of that for the denominators so, 4 x -1/4 on top and 1 the bottom will give me-1 on the bottom and 4 on the top so, now I have an eigenvector corresponding to the eigenvalue -3 .0731

So, when I find my homogeneous solution my homogeneous solution remember, I take each 1 of those eigenvalues and I make those the exponents for E T so, C1 E 2T knows my first eigenvalue and up of the eigenvector as a coefficient 11 x E 2T + C2 and my second eigenvector -14 x E the eigenvalue was - 3T .0756

So, that is my homogeneous solution to that system of differential equations .0796

I have not worried about the inhomogeneous part yet that is the GMT which will solve an example 2 just finding homogeneous solution right now .0804

Then for the fundamental matrix remember you take the 2 homogeneous solutions , you do not need the constants here so, I am in a take those 2 columns 11 x E 2Tis that is E 2T E 2T and then -14 x E -3T .0813

So, - E - 3T and 4 E-3T so, just take those solutions and I make those into the columns of a matrix and that is the fundamental matrix that were going to call sigh .0835

Were to use this E next example in example 2 to find a solution to the inhomogeneous system so, stick around for that but in meantime let me just recap how we found these solutions.0850

So, the key thing here is that even though we been given in inhomogeneous system examples ask us to solve the corresponding homogeneous system which means we really do not pay any attention to this GMT the inhomogeneous terms.0864

We just kind of throw them away for now these later in example 2 so, we write this in matrix form there is her matrix A and want to find the eigenvalues and eigenvectors of a .0879

So, I am subtracting R down the main diagonal and taking its determinant and I get a polynomial which I saw for R get to values for R each 1 of those I plug them back into A- RI .0890

Then I try to multiply it by a vector to get 0 and each time there I started with putting the 1 E Y coordinate and the just kind of figuring out what the x-coordinate should be.0904

So, that is where I got 11 for the first vector for second vector I got 1 and then a -1/4 E the top position.0916

And I did not like that because of the fractions so, because it is an eigenvector I can multiply it by scalars that is what I am doing here multiplied by 4 to get rid of the fraction and I got -14 .0925

Then our homogeneous solution just take the eigenvalues and you make them the exponents of E and then you take the eigenvectors and make those the coefficients and tack on a C1 and C2 to each part.0937

For the fundamental matrix what you do is you make each 1 of those homogeneous solutions into a column vector E individual column E fundamental matrix .0950

This E 2T E 2T that comes from the 11 E 2T and then - E- 3T and 4E - 3T that comes from that second homogeneous solution .0961

So, hang on to this matrix because we are going to use it to find the inhomogeneous solution in the next example .0975

So, example 2 being asked to solve this inhomogeneous system and what you notice here is that this is the same inhomogeneous system that we started working on an example 1.0983

So, this is the same as example 1 and we already started to working on that so, we already found our fundamental matrix back in example 1 and I am just going to copy what we got from there .0994

We found the eigenvalues the eigenvectors and so, our sigh from back from example 1 just a copy that matrix over from the previous slide was E 2T E 2T -E -3T and 4E - 3T .1015

So, we worked that out in example 1 if you did not just watch example 1 might want to go back and re-watch example 1 we will see where that matrix comes from. 1040

What I need to do remember for the key equation that were to solve for a variation of parameters is to find the derivatives of U1 U2 we find sigh inverse x GMT.1049

So, we need to find the inverse of sigh and part of finding the inverse of the matrix ABCD few will find the inverse of that matrix.1066

First thing we do is divide by the determinate AD - BC so, figures out what that is for this matrix right here the AD - BC for this matrix is E 2T x 4E - 3T .1079

Without before now E 2T x E - 3T is just E - T - - E - 3T x E 2T so, that is + because it is - - so, - in a - 3T x E 2T is E - T .1096

And so, we get 5 E the - T there and so, sigh inverse is 1 over the determinate so, 1 over 5 E - T and let me finish my formula for the determinate for the inverse here.1119

Remember U2 switch the D and A and you put - on the other entries - B and - C so, they fill that in here to switch the A and B entries so, 4E - 3T and E 2T.1139

Switch those and I negate the other 2 entries so, I got positive E - 3T and - E 2T.1158

And so, I want to multiply in that external factor and number 1 over E - T is the same as multiplying by E T so, I am multiplying by 1/5 E T and so, on get on the top left 4/5 x E T x E - 3T is E - 2T E - 3T so, 1/5 E - 3T x E T is E - 2T .1168

Remember and dividing by E - T that is the same as multiplying by either the T and then -1/5 E 2T x E T is E 3T and 1/5 E 2T x E 1/5 E 2T x E T is E 3T.1206

And I want to figure out sigh inverse G I am writing a little G everywhere here so, G and a copy E G which was the inhomogeneous terms from the original equation.1230

So, that is these 2 terms here got include the - sin so, E -2T and - 2ET so, if I work that through then I have to multiply these 2 matrices so, when I am going to get is 4/5 E - 4 T E - 4 T -2/5 E - 2T E T is E - T and then on the second row on multiplying the second row by the column on the right so, -1/5 E 3T E - 2T is E T now x - -2/5 E 3T x E T is E 4T .1249

So, that is my U' , what that means is that top row is U1′ and that bottom row is U2′ have to do is integrate each 1 of those to get the U1 and U2.1320

Slowly go ahead and fill that in U1 I integrate is the integral of U1′ DT so, me integrate that top row there 4/5 E - 4T the integral of that is just what you divide by - 4 so, it is -1/5 E - 4T .1338

And then I divide the next term by -1 because it is E - T so, I get +2/5 E - T and then U2 is the integral of U2′ DT so, -1/5 E T because I just integrates to itself .1368

Now I divide by 4 so, -2/5÷4 is -1/10 E 4T .1396

So, that is my U1 and U2 repackage them together as as into a column vector if you like and we will see what to do with next just a copy those over onto before I get rid of the slide .1406

Let me show you where each step came from so, were working with this general formula U′ is = sigh inverse G that is it absolute to a permanent formula for variation parameters but that sigh in that formula that Greek letter sigh comes from the 2 homogeneous solutions.1417

So, we found that back in example 1 you can look up an example 1 where this matrix came from so, there is that sigh to find the inverse of it we need to find its determinate first sigh found A/D - BC cross multiply their A/D - BC .1439

And I came out to be 5 E - T the inverse is 1 divided by that and then you mix up the entries in the middle this is D A and - C that is where each 1 of those came from where the original entries are A B C and D .1455

If you multiply that through then it is like dividing by E the - T but that is the same as multiplying by E T .2443 So, multiplying everything by 1/5 E T sets we got these entries here right each 1 of the exponents I bumped up by 1 so, -3 you - 3T x he is seeking the even - 2T and so, on there.1475

Now we want to multiply that by G from this formula up here so, this is our GMT and were getting that from the original set of differential equations so, multiplying this 2 by 2matrix by this column vector and it works out to be this not so, pleasant expression.1497

But each 1 of those tells us remember that U′ so, that U1′ and U2′ with integrate each 1 of those to get back to U1 U2.1518

So, I integrated the top row that vector on the right that is what I got when I integrated the bottom row that vector on the right that is what I got through.1530

Just to take this over and really use this on the next side would see how that works out.1539

So, here is the U1′ that we just worked out on the previous side there is the U2′ that we just worked out on the previous side and so, what to do is room multiply U1 x X1 + U2 x X2 .1544

So, just copy that down that is -1/5 of the little messy -1/5 E - 4T +2/5 E-T x my X1 is my first homogeneous solution E 2T E 2T now + -1/5 uses this is U2 that I am copying now used T -1/10 E 4T x my second homogeneous solution the X2 is - E - 3T and 4E - 3T .1564

So, multiply that through to get a little complicated on the top I see you have a -1/5 E -4+2 is - 2T and +2/5 E - T x E 2T is E T .1605

Go ahead and combine this with what I get from the second solution that the first row the second solution.1628

So, I have -1/5 x - here I get +1/5 E T E - 3T that gives me E - 2T and -1/10 x -1 gives me a +1/10 E4T E3T gives ET .1635

In the bottom row here I see have got while the first 2 terms of the same soldiers copy those quickly -1/5 E - 2T +2/5 E T.1659

Now the second on a second set of terms here I got -1/5 x E T so, x 4 E - 3T so, -4/5-4/5 E - 2T now -1/10×4 is +2/5 E 4T x E - 3T is E T.1672

So, let us see if that simplifies at all us looks at a should okay I see I have a -1/5 E - 2T and a 1/5 E - 2Tis nice and then I see have 2/5 E T +1/10 E T that is 4x +1/10/5/10 that is 1/2 E T .1706

On the bottom row I see I have got 2/5 looks like I may error down the bottom row that + should been a - that is coming from that -1/10 right there and that is nice that it works out that way because that means that 2/5 and that 2/5 E T they cancel each other out as I got -1/5 E - 2T and a -4/5 E - 2T.1726

Put those together you get -5/5 is - E - 2T.1754

What I just found there was my particular solution to the inhomogeneous system of the differential equations.1764

By the way there is feature here that is very common to variation parameters which is lots of cancellation at the end see how all these terms canceled at the end and that sort of all par for the course for variation parameters.1770

You expect that to happen and you often get very very complicated solutions that sort of gradually counts cancel down and turned the things that are fairly simple .1782

Let us put these together and get our general solution for general solution is the homogeneous solution + the particular solution .1793

And so, in this case our homogeneous solution we work this out back in example 1 is C1 x 11 x E 2T + C2 x -14 x E - 3T i am just copying this back from what we learned in example 1.1804

It had on the particular solution +1/2 E T and -E- 2T E - 2T so, that is my general solution we go ahead and box that often offer that is a formal solution there.1826

Alright let us recap how we found that a lot of the work was done the previous slide and we are over that we found the middle of mistake here and calling that U1' and U2′ .1851

This is actually U1 itself when we found and U2 itself on the previous side we found U1′ and U2′.1874

But we are integrated on the previous sites this is really U1 and U2 that we found here and so, we do is we take that U1 U2 and we multiply those buyer to homogeneous solutions.1881

The X1 there and the X2 so, after we multiply everything in and combine it into a single vector that a lot of cancellation which is kind of par for the course for variation parameters.1893

Simplifies down to this particular solution 1 half E T and - E - 2T and so, that is her particular solution to get the general solution we just add that onto the homogeneous solution which remember, we already found that much of the answer back in example 1.1907

So, homogeneous solution we found that in example 1 we just take that homogeneous solution and we add on the particular solution that we just found to find the general solution.1926

It is the end of the that example let us go ahead and try another 1.1937

So, example 3 were start out just by finding the eigenvalues and eigenvectors of the matrix 2 -5 1 -2 of course really use this later on E examples 4 and 5 directly solve inhomogeneous system of differential equations or just a start out by finding the eigenvalues and eigenvectors.1943

Remember , the way do that is you subtract Rs off the main diagonal so, that subtract those Rs C2 - R take the determinant now -2 - R - -5×1 +5 = 0 and so, I get R ² about a + to R - 2R and then -4+ 5 R ² +1 = 0 .1961

Now that does not factor of the real numbers you can use the quadratic formula we could just knows that this is R ² = -1 so, Rs equal to ± I .1993

So, got complex eigenvalues here so, let us go ahead and figure out the corresponding eigenvalues eigenvectors we should expect complex numbers to appear .2005

So, me plug-in articles I into A- RI and so, I get 2- I -5 and 1 -2 - I and I want that to multiply by an eigenvector and give me 0 .2015

Now my favourite trick here is to plug in 1 for the Y value and it was try to figure out what the X should be.2037

If I look at that second row there I see I get X -2 - I multiplied by 1 = 0 and so, X = 2 + I and so, my eigenvector is then 2+ I and 1 and so, they asked that the eigenvector corresponding to the eigenvalues R = I.2045

Now very nice property of matrices of real matrices with complex eigenvalues is that you only need to find 1 of the eigenvectors because what happens is the other eigenvalue hellos be a conjugate of the first eigenvalue and the other eigenvector will be a conjugate of the first eigenvector.2071

So, what that means is for R = - I don’t have to go through all this work I just know it can be the conjugate of the first eigenvectors can be 2- I and that is a 1 there.2089

Were conjugate just means you change a + π to a - π and so, those that is my Eigen value and corresponding eigenvector for the first 1 and then that is my second eigenvalue and corresponding eigenvector so, were done with that for the problem.2105

What we will do with the next problem is use these eigenvalues and eigenvectors to solve the differential of a system of differential equations.2126

Let me recap quickly how this worked out, we want to find A- RI want to find eigenvalues and eigenvectors so, you subtract R down the main diagonal and then cross multiply take the determinant get R ² = -1 .2135

You could use the quadratic formula that would work out very nicely on that but I did not think I needed to is I know how to solve R ² = -1 is just R = ± I.2152

Plug-in plugged that back in logic 1 of those articles I plugged it back into A- RI and so, I subtracted either on the main diagonal .2161

Looking for eigenvector and my favorite way of doing that is to put 1 in the y-coordinate and then just try to figure out what the X coordinates should be.2173

If you do not like doing that, you can solve this by linear combinations and free parameters the same way you learned linear algebra so, if you are not a call if you are not comfortable that then there is certainly other ways to do it.2182

But my way works okay and then when you look at the second equation we get X -2 - I x 1 = 0 as we solve X = 2+ I is I put plugged that 2+ back in for X and I got my eigenvalue and my eigenvector corresponding to I .2194

And the nice thing is that when you find the conjugate eigenvalue you can just take the conjugate eigenvector so, for articles - I just with the switch that + to a - and I get 2 - I and 1 for the corresponding eigenvector there.2215

So, in the next part of it examples 4 and 5, were going to do is use this matrix to solve for to solve the system of differential equations so, see how that plays out2233

So, in example 4 what were going to do is solve the following inhomogeneous system actually on the inhomogeneous system that were just being asked to solve the corresponding homogeneous system and then find the fundamental matrix and invert the matrix .2252

Were not really worry about the inhomogeneous system that were just going to try to resolve the homogeneous system and then ask the put these together and solve the the inhomogeneous system were put that off until example 5 .2272

You have a small typo here I wanted to make that a + cosine T a arithmetic works out better later so, were to change that to a + cosine T so, make little change there let us go ahead and work that out solve the homogeneous system first and that is nice because we have gotten started on this.2290

Remember in example 3 we already found the eigenvalues and eigenvectors of the coefficient matrix here 2 -5 and 1 -2 so, we found the eigenvalues and eigenvectors of the coefficient matrix there we remind you what 1 of them was.2312

The first 1 was R = I R = I, and the corresponding Eigen vector was 2 + I and 1 so, use that to solve the homogeneous system corresponding to this the inhomogeneous system that we were given.2330

So, what that means is remember the solution is E RT x the eigenvector V in this case the R is the complex number I and the eigenvector is 2 + I and 1 .2351

Remember, the way resolve such things we talked about this way the inner previous lecture where we had the complex eigenvalues so, if you do not remember this maybe go back and check that previous lecture relearned how to handle complex eigenvalues.2372

But the way to so, rt it out was we use this this identity E I θ = cosine θ + I x sin of θ .2385

So, that is what were going to invoke here this is E IT so, it is cosine T + I x sin of T and were to multiply that by 2+ I and 1.2398

And so, if I multiply that in I see I got 2 cosine T +2 I sine T and now the I terms + I cosine T I x I is - 1 a - sin of T and then on the bottom I just have 1x cosine T + I sine T 2419

And separate out a real part and a purely imaginary part for that so, separate out all the real terms and all the terms that have an I on them.2443

Look for the real terms I got it 2 cosine T - sin T and on the bottom of got a cosine T the all the purely imaginary terms I see I have a cosine T +2 sin T .2459

On the bottom sine T so, remember you do not ever have to worry about the second eigenvalue and eigenvector with these complex systems is you is you get it separated out like this then each 1 of these is 1 of your homogeneous solutions.2478

Slowly put these together and I will get the homogeneous set us the homogeneous solution my X homogeneous is C1 x that first solution there 2 cosine T - T over cosine T + C2 x that second solution cosine T +2 sin T over sin T.2501

So, that is my homogeneous solution we have elected that inhomogeneous part were just solving the homogeneous equations or not worrying about that that homogeneous part yet.2534

To find the fundamental matrix from the fundamental matrix, that just means you take the 2 homogeneous solutions and a packaging together as the 2 columns of a matrix so, we put those together.2549

My sigh will be understood for those 2 those 2 solutions as a columns of a matrix to cosine T - T and cosine of T down here and then cosine of T +2 sin T and sin of T down here .2562

So, that is my side and I will find the inverse of that we remind you of our formula for inverting a 2by 2 matrix ABCD inverse of a 2 x 2 matrix you divide by the determinant A D- BC .2586

Then you switch to the A and D supposed that a down there and you put - sins on the B C .2605

So, that means you got a find the determinant there so, to find the determinant size and sets part of my solution the determinant if I cross multiply there I get 2cosine sin can write the T 2 cosine sin - sin square that is multiplying that way and that if I multiply that way which I need to subtract all those terms.2620

- cosine ² - oh second got another 2 cosine sin and so, the 2 cosine sins cancel each other out and -sin ² - cosine ² is -1 so, sigh inverse is 1 over the determinants that is 1 of her -1 which is - .2652

Now just can go ahead and write my generic form D- B and a front for the might - C up here - C and A .2680

And so, that is - D positive B positive C and - A down here.2694

So, let us figure out what that is the context of this matrix number that is a ABC and D so, top left corner for sigh inverse of got a - D - sin T top right of got B so, that is just cosine T +2 sin T cosine T +2 sin T .2704

Bottom left of got a C that is cosine T and bottom right I got - A so, - A is the - of this so, that is sin of T positive sin T -2 cosine T.2725

So, that is my sigh inverse me go ahead rewrite the letter sigh inverse we will be using all this when we solve the inhomogeneous system in the next example.2746

But before we do that let me really use all these answers without going over them again so, they recap right now in show you where all these answers came from the first thing to notice here of the problem is asking us to solve the homogeneous system and so, what I am doing here is a really ignoring the inhomogeneous part of the of the system .2762

So, I am just kind of ignoring that for now looking at the homogeneous system so, that coefficient matrix which is 2 -5 1 -2 2 -5 1 -2 and we already found the eigenvalues and the eigenvectors there back in example 1.2787

In example 3 so, if you do not member how we got those eigenvalues and eigenvectors go back and check example 3 see how we got 2R = I for the eigenvalue 2and 2 + I over 1 for the eigenvector.2803

So, I plugged those in here E IT that is my eigenvalue right there that is the R and then there is the eigenvector and I expanded E IT using our old formula that we learned in the lecture on complex eigenvalues.2820

So, expanded that out of the cosine T + I sine T I multiplied that into the eigenvector multiplied into the top and multiplied into the bottom.2838

Gets kind of messy but we can so rt out the real terms of terms without an I and E terms with an I and remember what we learned about solutions for complex eigenvalues which do is you take each 1 of those and you make those each 1 of those columns 1 of your homogeneous solutions.2849

So, my general homogeneous solution is just those 2 columns multiplied by so, me arbitrary constant C1 and C2 and in fundamental matrix means that you just take those and make those the 2 columns of a square matrix.2866

So, there is my fundamental matrix and the last thing he asked us to do in this problem is to find the inverse of that so, I am using this well first I do find the determinant.2883

So, I cross multiplied here and it really worked out nicely, supply down into - 1 for my determinant so, I found the inverse that - right there that is really doing 1 over AD- BC that is why have that - on the outside .2894

Then I switched around the entries on the inside and when I multiply the - 1 and I got - DBC - A and I just read off my ABC and D from the fundamental matrix plugged those in and I got the inverse to the fundamental matrix .2909

So, use all this to solve the inhomogeneous system on the next slide see how that works out.2928

So, on example 5 here were to find a particular solution to the inhomogeneous system and remember there is a little typo there that should have been a + original had and solve it with a + there.2935

It is really messy if you work it out with a -sin solve it with a + and the way were going to attack this is really use the information that we had from the previous that previous examples.2951

A lot of the work has already been done in example 3 in example 4 so, me just remind you of what we figured out in example 3 in example 4.2966

We already figured out the fundamental matrix and we found its inverse and were ready to go to find the solutions to the inhomogeneous system.2975

So, what we learned at the beginning of this lecture is that you can find your derivatives of the U by doing the inverse of the fundamental matrix x G where G is the stuff right here.2986

That is our GMT it is the inhomogeneous terms of the differential equation so, we go ahead and write those out.3003

My U′ is there U′ is my inverse x G now inverse. We figured this out in example 4 so, we are not going to work it out from scratch again.3011

Example 4, that is what I am using right here to get my sigh inverse of just a copy that down from what we had is that the answer to example 4.3027

So, - sin T cosine T and then cosine T +2 sin T +2 sin T and sine T -2 cosine T that was the inverse aside.3038

Copy my GMT , that is coming straight from the differential equation here so, this cosine T & sin T and I need to multiply those through so, I see on the first row Ihave got - sinT x cosine T and then I got a cosine Tx sin T .3056

So, cancel it will just get 2sin ² T So, got that was I multiplied this row by this column and I saw that I had the 2 terms canceling the term of cosine T sin T in as occurring with the - and with the positives that is why cancel out those out right away.3081

Now we multiply the second row by that same columns I see about cosine ² + sin² set the 1 -2 sin T cosine T 2 sin T cosine T.3103

And I think that remember that is U's what I am going to do is use so, me trigger metric identities here geometric identity and many used to make it easier to integrate is +sin ² of T = 1/2×1 - cosine 2T .3122

That is a trigonometric identity and if you do not remember that we got so, me lectures on trigonometry here on that educator and we also, use that pretty heavily in calculus 2 lectures here on educator .3141

Calculus 2 learn how to integrate trigonometric functions and 1 of the tricks there was when you have a sin ² cosine ² you want to use this double angle identity so, that is what I am kind of recalling that from 2.3154

And so, if I write this out I get a get well 2 sin ² T would just be 1 - cosine 2T and then I saw the bottom I have 1 -2 sin T cosine T .3171

So, I really want to think of those as that U′ that U1 U2′ so, U1 is the integral of that first row 1 - cosine 2T DT and so, by integrate that I just get T.3189

Now the integral cosine assine but since it is cosine 2T it is 1 half sin of 2T.3211

You do not have to include the constant when youare finding this U1 U2 if you did include the constant it would just give you more multiples of the homogeneous solution so, you do not have to worry about the constant here.3219

The U2 work a little more with U1 I am going to use another trig identity there which is that the sin of 2T = 2 sin T x cosine T and so, T - ½ 2T is just T -1/2 of what I just wrote.3234

So, T - sin T cosine T to go to be a little easier to work with later on, go ahead and find U2 that is the integral this was U2′ the top row was U1′ so, to integrate U2′ to get U2 .3260

So, 1-2 sin T cosine T that was a lot of ways you could integrate this but I think the 1 that is going to work the best is to do a little substitution here for cosine T and I like using U substitution but some are using the variable U for something else.3278

I am in a use of different variable interviews W W = cosine T and so, DW = - sin T DT and so that will give me if I plug those and now do maybe integral of + 2W because the - is included E DW + 2W DW .3299

So, the integral 2W is just W ² so, that it be very easy so, when I integrate this the 1 just integrates to T the 2 sin T cosine T integrates to W ² so, that is + cosine ² T and so, I have got my U2.3325

Were to figure out how to use that on the next slide but let me just quickly recap what we did here we already found sigh inverse back in example 4 so, our sigh inverse that comes from example 4 if you do not remember how he got that just go back and read over example 4 you will see where it came from.3348

And our G is coming right here from the inhomogeneous terms in the differential equation so, plug those in right there I multiplied to the 2 rows there and there was so, me cancellation going on and a simple fight a bit into 2 sin ² and 1-2 sin T cosine T .3364

Now my sin ² T I use this old trigonometric identity a lot and calculus 2 to convert that into 1 - cosine of 2T .3388

So, then what I have here is U1′ and U2′ and I integrated each 1 of those little substitution here turn it into his ½ of 2T and then I used this old trigonometric identity to simplify that into sin T x cosine of T .3398

Then the second 1 I use a little I wanted to call you substitution but so, me are using you for something else I call it W so, W = cosine T and then DW = - sin T .3417

So, had 2 W and that integrates to W ² so, we get T + cosine ² T that T came from that 1 right there and then need the rest of the game in a cosine ² T .3432

Since my U1 U2 keep going with that on the next slide so, in our next slide we have our U1 here that is actually not quite what we had for U so, on the there is there is a couple extra terms your little typo here slowly fix that quickly.3445

The U that we had was T - sin T cosine T and bottom part erase that in fix that quickly in the bottom we had 2+ cosine ² T .3471

So, that is the answers that we got from the previous sides go back in and wants at previous side if you do not remember where they came from and so, now let me show you how you use these to write down the completes solutions of the problem.3492

So, far particular solution is let us see it is a U 1 x our X1 + U 2 xX2 .3506

So, that is can be a little elaborate and complicated might right that out there U1 is T - sinT cosine T and X 1 we get from this first column of the matrix so, that is 2cosine T - sin T and cosine T in bottom.3527

Now + U2 so, that is T + cosine ² T + cosine ² T and X2 to get that from the second column of the fundamental matrix so, cosine T +2 sin T & sin T on the bottom .3557

Now this gets to be a little ugly incompetent but it is going to simplify nicely after we plow through all the mess so, let me out work through that first row here I see about 2 T cosine T - T sin T -2 sin T cosine ² T + sin ² T cosine T .3583

And to go ahead and add on the first row of the second term there so, + T cosine T + 2T sin T + a cosine 2T +2 cosine ² T sin T .3611

So, very long and messy but I am hoping there is going to be so, me good Cancellation coming here.3637

In the bottom we get T cosine T - sin T cosine ² T cosine ² T + T sin T + cosine ² T sin T 3642

So, I am really hoping so, me will simplify and it is customary when the new variation parameters to get very long complicated solutions and then they do somewhat simplified.3661

So, if I look at this top term here I see I have got 2T cosine T and I got another T cosine T so, and put those together get 3T cosine T now - T sin T + 2T sin T so, - T + 2T so, just + 1 after all that simplifies now -2 sin T cosine ² T and +2 cosine ² T sin T .3672

Those 2 terms just cancel each other out that is a really nice and I see got 2 other terms here + a sin ² T cosine T forgot my T cosine T + cosine cube T .3715

Now the bottom I see I got T cosine T + T sin T and then I got a -sin T cosine ² T + cosine ² decided those to cancel so, really did cancel quite nicely and it actually is no work out even better because if you look at these 2 terms I can factor out a cosine T x sin ² T + cosine ² T .3733

Of course cosine ² + ² is just 1 so, those 2 terms R to combining just give me a cosine T and so, the thing we will do is factor out at T whatever terms I cannot get T x 3 cosine T + sin T .3762

And then the bottom I got cosine T + sinT and I still have this cosine T so, let me not forget that is just E top so, I am in a factor that out on the right get a cosine T and other say I have 10 because I had that term only appears on the top and everything else canceled on the bottom.3786

I think on this 1 were just asked to find a particular solution so, that is my particular solution so, let me recap and show you all the steps that went into that.3811

This sigh first of all came from solving the homogeneous system we did that back in example 4is that is where we found that sigh that is example for coming in and being useful here.3826

Watch example 4 recently so, back and check example 4 you will see exactly where that side came from .3840

These 2 entries for U that was U1 that was the U2 that we found on the previous page originally wrote it wrong in my in my sigh so, I just corrected it right there that that is the T + cosine ² T which we found the previous page .3848

Then our general form for the particular solution is U1 x X1 + U2 x X2 so, copied my U1 down there copied my X 1 that was the first column of sigh there is my U2 and then I copy my X 2 that is the second column of sigh X2.3866

So, we always have this form U1 X1 + U2 X2 every time you variation parameters you got that form and then multiplying this through the first row that gave me these first f4R terms there.3889

Multiplying this through the first row gave me the next 4 terms of very ugly expression there.3902

The multiplying this through the second row gave me those 2 terms and then multiplying this to the second row gave me those 2 terms.3912

So, get this really horrible expression but it is nice a lot of things cancel the these 2 terms cancel this 1 cancels with this 1 this term combines with this term the T sin T combines with the 2T sin T .3923

So, we end up with 3T cosine T T sin T and these other 2 terms sin ² T cosine T + cosine QT and those 2 we can factor out and simplify that 1 just down into cosine Tof this whole thing is turned in a cosine T .3943

In the bottom we get terms canceling those big terms cancel we just get T cosine T + T sin T and then what I see now basically I am done I just want to factor the T out make him look a little nicer so, I factored out that T from the first 4 terms factor that out here .3963

That left me with a 3 cosine T + T and a cosine T + T and I still had that 1 terminal cosine T can factor T out of that side to write it separately so, there it is 10 x cosine T .3982

Sets the particular solution to that inhomogeneous system that we start with way back in example 4 then had again in example 5 .3998

If you want to find the general solution it would really be no more work because what you do is you would just take the homogeneous solution and then add on the particular solution.4006

We partly found both of those copy them again because it is kind along and there is really nothing to be gained here but we found the homogeneous solution in example 4 .4019

Then this particular solution is what we just found right here so, if you want to you can just put those together and get the general solution to that inhomogeneous system of differential equations.4030

So, that is the end of our lecture on solving inhomogeneous systems of differential equations .4043

Remember we had 2 methods there was a bit undetermined coefficients that was the previous lecture and then variation parameters that was this lecture .4051

So, 2 completely different ways of finding this particular solution but then they both work and they both end up giving you solutions to inhomogeneous systems of differential equations.4059

That actually wraps of this chapter on systems of differential equations.4072

Our next chapter is going to be on numerical method so, totally different stuff I hope you will stick around and watch that chapter as well.4076

In the meantime, you are watching the differential equation series here on educator.com. My name is Will Murray and thank you very much for watching, bye bye.4083

Hi and welcome back to educator.com, my name is Will Murray.0000

We are covering differential equations, today we are going to study numerical techniques and in particular were going to cover Euler method.0002

Now, we do have another lecture on Euler equations and that is a totally different topic.0010

So, if that is what you are looking for If you are looking from Euler equations then you do not want to be watching this lecture.0016

But you can go over and see another separate lecture on Euler equations so, this is Euler’s method and only go ahead and start explaining what that is all about.0022

So, Euler’s method is a way to find numerical suits approximations for initial value problems that we cannot solve analytically .0031

We have all these techniques to solve differential equations analytically variation parameters and undetermined coefficients of linear equations and separable equations penalties analytic techniques where we can solve sort of using algebraic techniques and find exact solutions.0042

When you use Euler’s method, is when you cannot use those analytic solutions and so, you have to use approximation techniques so, we are going to be doing a lot of computing here there is not a whole lot of theory involved in Euler’s method.0059

But there is a lot of just plain numbers into equations and calculate see how it goes.0072

It is based on drawing lines along slopes in a direction field so, the idea is that you will start at some initial point Let me start at some initial point here and then you will kind of follow the slope along for a while .0078

Then you recalibrate you recheck your slope and follow that slope along for a while and then you recalibrate and so, on and keep doing so, that is kind of the basic idea of Euler’s method is just under there is not not a lot of theory involved it is just a matter of walking along in the solution space and rechecking your slope at every step.0091

So, they show you how the details of that work out what you do is you start at with an initial value problem so, those who always be in the form Y′ = some function of T in Y.0115

See some example of this you will get the hang of it and then you always have an initial point Y t not = Y not .0128

So, what you do is you start at that initial point T not, Y not and then you are going to move over in steps and so, in order to do that you have to know, what your step sizes that is what this value Hs.0135

That is usually given in the context of the problems so, the problem will say solve this such and such a differential equation using Euler’s method with step size 0.1 or 0.05 or 0.5 or something like that.0159

So, the step sizes H that is usually given to you in the context of the problem and what you do is you try to go over from T not Y not and you try to find the next point T1Y1.0167

And from there you find the next point T2Y2and here the equations that she used to do it to get to TN + 1 you just take TN and you add on the step size H .0179

So, you just stepping over a value of H each time it is going over and regular steps in the T direction2get the new Y value you start with the old Y value Y at and you add on H x now, this F TN YN this F TN YN that comes from the differential equation .0195

So, you go back 2 the differential equation and you plug the current value of TN YN into the differential equation and you evaluate that and then multiply by H and that tells you how far your stepping vertically.0220

So, really what this is doing here is it sort of the rate x time kind of thing the H is the time the change in time and the F is the rate because that is the same as Y′ .0234

This sort of keeps track of horizontal steps this is a your taking a horizontal step over from TN to TN +1 and then here your taking a vertical step up from YN to YN +1 .0249

Now, usually in the context of the problem they will tell you a certain value of T where they want you to approximate Y of T and so, what you do is you keep making steps over until you get to the value of T that you looking for so, let us try that out and see how that works in practice.0267

So, your first example we use Euler’s method was step size 0.1 to estimate Y is 0.4 in the initial value problem Y′ = 1+ T - Y and Y0 = 1 .0286

So, let us see how that plays out what this initial condition this Y0 = 1 that tells us that T0 = 0 because of the value of T there and Y0 = 1 that is the initial value of Y that we are given.0302

Were also, given this is F of T, Y that is always the function that Y′ = some function of T, Y and let us remember basic equations for Euler’s method all right them up here in the corner.0321

TN +1 is always equal to TN + H TN + H is the step size and YN + 1 is always equal to Y N+ the vertical step is H x F of TN YN.0337

So, let us start figuring those out the F here the F of T0Y0 F of T0Y0 is F of 01 and now, since F of T Y is 1+ T - Y that is 1 +0-1 which is 0 and so, are were to find out what T1 is T1 is just 0+ H is is T0 + H all right that is T0 + H which is T0 was 0 H is 0.1 .0357

Our Y1 following the formula there is Y N so, Y0 + H x now, F01 is just we are figured out is 0 so, that is just Y0 which we already said was 1.0408

So, we got our T1 right there 0.1 and our Y1 is 1 and now, let us just Keep repeating that procedure F of T1Y1 is F of 0.1, 1 which again F of T Y is 1+ T - Y so, it is 1+0.1-1 which is 0.1 .0424

And so, our T2 is T1 + H so, T1 was 0.1 Hs 0.132 is 0.2 and our Y2 is Y1 + H x 0.1 getting the 0.1 from this 0.1 here so, Y 1 is 1 + now, Hs 0.1 so, 0.15 0.1 is 0.01 which is 1.01.0452

So, that is our T2and there is our Y2 and order to keep going until we get to the T value that were asked for which is that 1 right there like you going to get2T equal 0.4 so, looks like we got2more steps here .0492

So, we go ahead and calculate F of T2Y2which is F of 0.2 and 1.01 which is 1+ T - Y so, 1+0.2-1.01 and that is = 1.2-1.01 is 0.19 and now, we can write down T3is T2 + H which is 0.3 and Y3 is Y2 + H x that value we just we just figured out the F value was 0.19 .0510

Now, the Y2 was 1.01+0.19 x H H is 0.1 so, that is the 0.019 and so, we had those together and we get 1.01 1.019 is 1.029 so, we found our T3 is 0.3 our Y3 is 1.029 3 quarters the way there.0571

Lets find our F of T3 Y3 is F of 0.3 and 1.029 which is again F is 1+ T - Y we get that from the differential equation 1+0.3-1.029 and if we simplify that down we get 1.3-1.029 that is 0.271 .3-.029 .0610

So, now, we can find our T 4 is T3 + H which is 0.4 and our Y 4 = Y3 + H x our F value from above which is .271 which is now, Y3 was 1.029+ now, H is 0.1 so, 0.1×.271 is .0271 and so, if we add up 1.029 and 1 point 0 .271 we get 1 .05610653

Let me summarize here our T 4 Y 4is 0.4 and 1.0561 and so, that value right there is our estimate of the Y value when T is .04 .0706

So, that is our estimate for Y 0.4according to Euler’s method so, they recap and show you what I did here a sort of the basic equations for Euler’s method the TN + 1 is always TN + H YN + 1 is YN + H x F of TN YN where we say F we using this function right here from the differential equation.0728

So, start out T0Y0 comes from the initial condition T00Y0 is 1 and then I plug those into F and it just gives me 0 on the first step I always find my T1 by adding on an 8 sided T1 0.1 Y 1 from that equation is Y0 + H.0753

Now, that is that 0 is going in there just turns out 2 be 1 again soon I got T1 and Y 1 of plug those back in F I got 1 + T - Y and that goes down the 0.1 so, T2 I always think does go through all the T 0.1 0.2 0.3 and 0.4 those are all the T the Y at each step I take the previous Y + H x the F values that .01 is going in there.0776

We found the F value at the next step we got .19 and so, we plug that in to find the next Y3 found the F value there to get .271 and so, that plug-in there to get Y4 that simplify down to 1.0561 and so, I stop there because I was asked to find Y of .0 0.4 so, I stop there because I have got to achieve value of 0.4 and I take my corresponding Y value as 1.0561 .0814

I offer that as my estimate of Y of 0.4 remember Euler’s method is always an estimation game it never actually tells you an exact answer.0850

So, what were going to do in the next problem keep this problem in your hand and keep our answer in your head because what we are going to do in the next problem is we are going to solve this differential equation analytically and were going to find the true solution and compare it with a solution that we came up with here using Euler’s method 2 approximate.0862

So, an example 2 we will solve the initial value problem Y′ = 1+ T - Y and Y 0 equal 1 were to solid analytically.0882

And order to compute Y of 0.4 now, this is the same problem that we solve using Euler’s method in example 1.0895

So, we are going to compare our exact answer that were defined in this example to our approximate answer that we found for example 1 were to see how close they are 0903

So, let us work this 1 out we got there we got Y′ = 1+ T - Y will bring the Y on the other side get Y′ + Y = 1+ T .0914

Now, this is a linear differential equation and back to the very beginning of the course and remember my techniques for solving linear differential equations so, if you do not remember how to solve linear differential equations is been quite a long time since we talked about that.0928

But you might want to go back and re-watch that lecture on linear differential equations because we get a whole lecture together a linear differential equations and so, that is most of the something fairly familiar to you by now, .0947

But it is a little rusty just go back and check out the lecture and you will see how to solve it just write down the name linear differential equations.0959

And so, the technique there is to use this thing called the integrating factor where you do I T = E to the integral of P T DT .0971

And the P is whatever the coefficient of the Y is so, this case is just E of the integral of 1 DT which is ET and so, what you do is you multiply both sides by that integrating factor.0988

So, multiply everything Y′ x E T + Y x E T is equal to 1 x E T + T x E T .1004

The point of doing that is that makes the left-hand side into the derivative of Y E T using the product roll again this is all coming straight out of the lecture on linear differential equations 1712 so, go back and check that out if your rusty on that E T + T x E T and so, we can integrate both sides and on the left will just get Y E T on the right will get E T 1019

Now, the integral of T E T you really have to use parts and I may use a little shorthand technique for parts T x ET shorthand technique is right derivatives down the left-hand side and integrals down the right-hand side.1051

Then multiply down these diagonal lines with an old with alternating sins and so, I get + T E T - E T + C1070

That was really integration by parts and that was something that we covered here in the lectures on educator.com in the calculus to section so, you might want to check the lectures from calculus 2 on how to integrate by parts.1084

We have lecture there on integration by parts and if you do not remember how to do that just go back and watch those lectures and you will see you get a refresher course and integration by parts.1096

So, meantime we got Y ET = R T + T R T - R T + C cancel those as I get Y E T = T E T + C and now, I am going to divide both sides by E T so, get Y = just T by itself + now, if I divide by E T that is the same as multiplying by E -T have to keep track of the C at this point .1107

Now, 2 figure out what the C and when he is my initial condition Y 0 = 1 Y 0 is just equal to 0+ C E 0 which is C and that is both equal 1 so, my C = 1 and play that back into my equation.1142

Y = T + C is just 1 E - T .1163

So, that is my analytic solution to my differential equation and I want to plug-in 0.4 and see what happens there so, Y is 0.4 = 0.4+ E -0.4 .1169

That something that I want to figure out using my calculator so, I am going to go 0.4+ E -0.4 and it tells me it is approximately = 1.703.1188

So, that is what we get solving the equation analytically and let me just remind you what we got using Euler’s method, we use Euler’s method member that was an approximate solution without solving the equation analytically.1213

Euler’s method we did this in example 1 so, Just watch example 1 go back and check that out we found that Y is 0.4 was approximately equal to 1.0561 1228

That is what Euler’s method gave us so, he left out a 0 when I found my exact solution it was 1.0703 for my calculator told me and I just did not write down that last 0.1247

Here is the solution got by doing Euler’s method and that was example 1 if you have a just watch that.1269

That is where that answer came from and so, we can see how good our approximation was using Euler’s method it was out 1.05 instead of 1.07 which is pretty close it is not perfect but that is the whole price of using approximation techniques is you would not get an exact answer.1279

So, the exact answer would really be 1.0703 but in order to do that we had to solve the equation analytically.1297

Let me show you the various steps involved here first I took the differential equation I move the Y over 2 the other side and then that is a linear differential equation.1307

So, using the techniques we learned in our early lecture on linear differential equations I found the integrating factor in the integral of PNT where PNT here is just 1 there so, we just get E T multiply that by both sides.1317

That turns a left-hand side into the derivative of a product using the product roll so, we integrate both sides we just get Y E T on the left on the right we had integrate by parts T E T ahead integrate by parts course the inability the T by itself is just E T .1332

So, when I integrated by parts again using what we learning calculus 2 we get T E T - E T 2 terms cancel we just get T E T + C .1351

You do have to include the constant when you do the integration is kind of a very T lesson of differential equations and so, will resolve for Y were multiplying both sides by E - T so, get T + C E - T .1364

Then we use the initial condition T is 0 on Y is 1 I played in T0 I set it equal to 1 and I get C = 1 pretty convenient plug that back in to the general solution here and that is how I get the specific solutions the initial value problem Y = T + E-T .1378

That is I compute Y = 0.4 so, plugged in T is equal 0.4 for T plug that in and this is a something I worked out my calculator 1.0703 and I compare that with what we found in example 1 for Euler’s method .1398

Using Euler’s method to look at the same differential equation Y and 0.4 back there we calculated 1.0561 so, we really got 1.05 instead of the true answer would be 1.07.1419

That shows you the Euler’s method is pretty close but it does not give you a perfect solution so, let us keep going here with example 3.1431

Right with example 3 we are going to use Euler’s method was step size H = .01 and will estimate Y .0 for the initial value problem Y′ = T ² + Y ² and Y 0s equal to 1.1442

So, let me set up our starting point here we got T0 = 0 and Y 0 =1 getting those from the initial condition and down than I just remind ourselves of the Euler’s method equations T N+ 1 is always TN + H and Y N+ 1 is always YN + H x F of TN Y N .1459

And that were talking about is this function right here T ² + Y ² so, that is what we are going to use to compute the T ² + Y ² so, my T1 is 0+ H which is 0.1 .1494

My Y 1 is Y0 + H x now, T ² + Y ² is 0 ² +1 ² so, Y0 was 1 and + H x while 0 ² +1 ² is just 1 so, H is 0.1 and got that from over here and so, that is 1.1 and I have a T-1 and I have a Y1 Let us go and find T2 .1513

0+ H is 0.2 my Y2 is Y 1 + H x the function of T-1 and Y 1 so, the function here is T ² + Y ² so, that isis 0.1 ² that is from the T-1 +1.1 ².1553

Now, my Y 1 was 1.1 my H is 0.1 and .1 ² is .01+1.1 ² is 1.21 and so, that gives me 1.1+0.1 x looks like 1.22 and so, that is 1.1+.122 .1579

If we add those together were going to get 1.222 so, now, I have a T2 and I have a Y2 .1615

Keep going to get 2 more steps because were shooting for Y have 0.4 so, T3 is C is a small mistake here when I said T been T-1 + H I do get the right answer was 0.1+0.1 this should be T2+ H which is 0.2+.1 so, .3 .1630

The time stamps are very easy to keep track over to stepping along by 0.1 each time the Y steps are harder Y3 = Y2+ H x F of T2 and Y2 so, that 0.2 was T2+1.222 ², Y2 was 1.222+0.1 and start using a calculator here.1656

0.2 ² is .04 but 1.222 ² so, 1.222 ² is 1.493 and about 1.222+0.1×.04+1.49 is 1.53 and 0.1×1.53 is .153 and so, if I add that to 1.222 I get 1.375 for my Y3.1692

So, T3 and I got Y3 just 1 more step to go T 4 is T3 + H is 0.4 my Y4 is Y3 + H x F of F of T3 and Y3is a T3 was 0.3 and our function value is T ² + Y ² so, 1.375 ² and let me keep going in the new column here.1740

My Y3 was 1.375 my H is 0.1 and 0.3 ² is .09 1.375 ² we get 1.89 and .906 go ahead and add .092that +.09 and then I multiplied by .11784

The rest of this on my calculator and then add 1.375 to that and it looks like it also, provides down to 1.573 so, that was my Y4 remember.1819

And appointed that is the best my estimate for Y of .4 so, I got my T4 and my Y4 summarize that T4 Y 4 is 0.4 and 1.573 53 but 573 so, that right there is my estimate for Y of 0.4.1841

So, let us go back over that make sure that everybody understands all the steps.1872

We start out with G0 = 0Y0 = 1 that came straight from the initial condition and we started stepping through this procedure following the initial equations for Euler’s method.1881

TN + 1 is always TN + H so, my H here is .1 that is how I hate my steps go .1 .2 .3 and .4 that shows you how the T develop the wiser more collocated following this formula it is always YN + H x F of TN YN and this is the F right here.1894

T ² + Y ² so, that is why 2 find Y 1 I took Y0 + H x now, that is T0 NY 0 there that is T0 and Y0 and when I found Y2I did Y 1 + H x that is T-1 and that is Y 1 down here to find Y3 it is Y2+ H x T2 and Y2.1918

And finally to find Y 4 its Y3 + H of that is T3 and that is Y 3, crunch all these numbers we end up with Y4 = 1.573 and I know, that that is where I want to stop because I got into my T value of 0.4 which is what I was asked estimate in the initial example.1949

And so, my Y value 1.573 is my estimate there so, example 4 we use Euler’s method was step size H = .0 or 0.3 and were to estimate Y is 0.6 in the initial value problem Y′ = T - Y and Y0 = 1 suddenly remind you of the equations that we always use for Euler’s method .1972

TN + 1 = TN + H stepping along by values of H and YN +1 = YN + H x F of TN YN.2004

Now, that F is just whatever you get from the differential equation that is F of TY right there so, the way you start is with the initial condition that your givens and that is Y 0 = 1 so, my T0 = 0 and my Y0 = 1 .2020

And I will start stepping through the algorithm but my step sizes H so, T1 = T0 + H which is 0+.3 so, that 0.3 my Y1 = Y0 + H x F of TN YN so, that is T0 - Y0 .2039

I am getting that from here T - Y so, that= now, where to my Y0 there is its 1+0.3 now, T0 - Y0 is 0-1 so, this is 1 +.3 x -1 1-.3 is 0.7 . 2065

So, now, I found T1 and 0.3 and Y 1 is 0.792 is T1 + H which is 0.3+0.3=0.6 and my Y2is Y 1 + H x F of T-1 Y1 so, T-1 - Y1 .2088

So, my Y 1 was 0.7+ where to my H go 0.3 now, T1 - Y1 got my T1 was 0.3 - Y 1 was 0.7 and so, this is -0.4 if you must buy that by .3 get -0.12 so, got 0.7-0.12 which is 0.58 .2119

And I'm going to stop there in the reasonable. There is because I was asked to find Y of a 0.6 and look here we got T2 is 0.6 so, I may use my Y2that value as my estimation.2154

So, to summarize there the T2 Y2is 0.6 and then my Y2 was 0.58 and so, I am in a hang on to that 0.58 and offer that as my approximation for Y of 0.6 .2177

So, let me show you the steps again there we sort out T0Y0 came from the initial condition there and we started running through this condition these these equations. 2200

The TN +1 steps by point 3H time steps by H each time so, we go from 0 2 .3 2 .6 and then to find the YN +1 we do YN + H x F of TN YN, and our F of T Y is here 2 my T - Y .3710 Since Y CT 0 - Y0 here and T-1 - Y1 here so, we figured out Y 1 was 0.7 Y2 is 0.58 and I stop there because I got into my T value of 0.6 which was the desired 1 from the problem.2212

So, may offer that 0.58 my last Y value as my estimation for Y0 .6 remember this is just an estimate because we are solving this thing numerically weird solving it analytically.2248

What we are going to do in the next example is to solve the same initial value problem analytically will see how close our answer is to the answer we just figured out.2262

So, in example 5 here we will solve the initial value problem Y′ = T - Y and Y0 = 1 were solid analytically using what we learned much earlier in this lecture series. 2279

Remember, calculate Y0 .6 and compare with the answer with the result given by Euler’s method we figured that out in example 4 so, let us go ahead and start solving this .2289

I am not right Y′ + Y = T and now, this is a linear differential equation and we learned how to solve linear differential equations in 1 of the very early lectures here in the differential equation series on educator.com .2301

What you do is you find the integrating factor = E to the integral of P of T DT where your P is whatever the coefficient of Y is right there in this case that is just 1 so, in the integral of 1 DT is just E T .2322

And so, you multiply both sides by that integrating factor so, we get Y′ ET + Y ET =2E T .2341

And then the point of that is what we learned in electron linear differential equations is that the left-hand side is the derivative of Y E T using the product role right-hand side of T E T so, we can integrate both sides.2354

Left will just get Y ET because were integrating a derivative and on the right now, many use parts any as integration by parts which is something we learned about in the calculus2lectures here on educator.com is 1 of the early results there.2373

So, in the integrate T E T using integration by parts is using tabular integration it is kind of a shorthand to solve certain integration by parts problems so, writing derivatives on the left integrals on the right and then I am going to put my alternating sins + and - sin it T E T - in the T .2390

And I will is have to attach a constant at the moment when I do the integration and so, I am going to solve for Y there divide sides by E T so, get Y = T-1+ C by E T that is the same as CE -T .2416

And that is my general solution now, I am going to use the initial condition Y0 = 1 so, 0 = 0-1+ C 0 and that is supposed 2 be equal 1 . 2435

And so, what I see quickly here is that C must be 2 because E 0 is 1 and all play that back in my general solution Y is equal 2 T -1+2 in the negative T.2451

And so, that is my solution to the initial value problem Y = T -1+2 in the negative T and I wanted to compute Y of 0.6 so, the plug-in 0.6 for T 0.6 is 0.6-1+2 E -0.6 which it looks like that would be -0.4+2E of the -0.6 .2467

Indefinitely is my calculate for that so, -.4+ 2E to the -.6 and that gives me well approximately if I just take the first 2 decimal places 0.6976 as an answer.2504

And that is really an exact well that is very close to being an exact answer let me remind you what we got using Euler’s method so, we can see how accurate are Euler’s method answer was.2529

In Euler’s by using Euler’s method we did this in example 4 so, you can go back and check out example 4 if you are not sure where this is coming from we got Y of 0.6 .2545

Our estimation from there was 0.58 so, that is the estimation that Euler’s method gave us Y 0.6 0.58 the true solution according for analytic techniques is 0.6976 .2558

So, we got something fairly close but obviously not completely accurate so, Let me remind you what we did with this we saw a linear differential equations were many use my solution techniques for linear differential equations and that is to use the integrating factor.2579

This is coming from the lecture very early on in this series a linear differential equations into E integral of P of T were PT is the coefficient of Y so, here is just E integral of 1 so, that is E T .2596

Multiply that by both sides that turns the left-hand side into the derivative of our products are really kind of explaining the product role there so, if we integrated we just get that original product back Y E T .2609

On the right-hand side we had to use integration by parts to get E T - E T + C and so, solving for Y that means multiplying both sides by E -T .2622

This is what we get for general solution that we as the initial condition Y0 = 1, plug that back in and we get C = 2 so, I plugged that back into the general solution and that is how I got my specific solution here T -1+2 E-T and out to estimate Y is 0.6 .2635

Plug that in for T and I get something that I certainly would not want to do by hand but on a calculator.2658

That tells me it is 0.6976, and want to compare that with my answer from Euler’s method we worked out the answer from Euler’s method in example 4 set for this number 0.5 A comes from.2667

Worked on that just above an example 4 so, that is the Euler’s method answer the true answer is 0.6976 and so, you can see that 0.58 versus 0.69 we were fairly close with Euler’s method we were not completely accurate.2680

In the next lecture here on not the differential equations lecture series, we will talk about the Dakota method and will solve the same problem receive to get something much more accurate and so, will compare that answer whether Euler’s method answer will see that for the price of doing a little more computation, we can get a much more accurate solution. 2695

So, that is the end of our lecture here on Euler’s method as part of the differential equations lecture series on educator.com. My name is Will Murray, thanks for watching, bye bye.2718

Hi and welcome back to the differential equations lectures here on educator.com .0000

My name is well Murray, and we are studying numerical techniques right now, we are about to learn about the Runge-Kutta method.0004

Now, this is also known as the improved Euler method so, you might also, see it called the improved Euler method this is to be distinguished from the sort of basic Euler method and we had another whole lecture on the basic Euler method.0012

That is the previous lecture here on educator.com so, if you are looking for the basic Euler method and that is the previous lecture if you are looking for the improved Euler method, that is this lecture right here keep watching so, find out what that is all about.0027

The Runge-Kutta method, also known as the improved Euler method is a way to find numerical approximations for initial value problems that we cannot solve analytically.0043

So, the idea here is that we have some differential equation and some initial condition and we cannot solve it analytically so, we use these techniques of numerical approximation instead.0056

The Runge-Kutta method is more sophisticated than Euler’s method that is what it sometimes called the improved Euler method, it is actually the method that is built into the programming packages.0069

If you use commercial software to solve differential equations so, it is things like Mathematics or Matlab, then the when you solve differential equations numerically on those software packages still be running some version of the Runge-Kutta method in the background.0080

So, we are going to learn the order to run Dakota method , that wins it is already pretty complicated what the F professional software packages uses something called the order for even more sophisticated methods and that is a little too collocated for us to go into by hand.0100

But we will learn the order to run Dakota method annual see him get a flavor of how it works.0120

So, let us check that out and get down systems actual algorithms so, you start with it an initial value problem you will have Y′ of T = some function of T and Y .0126

They also, have an initial value Y T = Y not and shall choose a step size and usually when you are studying differential equations the step size will be given to you it would not be your own choice to choose the step size.0140

So, that will usually be given of course in real life you get to choose your own step size depending on how accurate of an answer you want and how much work you want to do.0155

So, you start at T0Y0 and you make iterative steps to me fill in this the T N+ 1 it is just the same as it was for the Euler method it is just TN +8 so, just stepping over in horizontal steps of size H 0164

Y MT TN +1 so, this is Y and + 1 Y N + 1 is the same as Y TN +1 now, this is much more complicated than it was for Euler’s method so, let me show the steps for this carefully.0184

It is Y N+ now, H instead of just H x F of TN YN that was the Euler’s method formula its H x K-1 + K2 over 2 see what they did this is an average because it is and what is it an average out.0205

Let me explain what these that this K-1 and K2 are so, K-1 is just F of T and Y so, that is what we had that is the same as it was for Euler’s method the K2 is different K2 is F of now, for TU plug-in TN + H and for Y you plug-in Y and + H x K1.0223

So, that is quite a bit more complicated so, what will do each time is will calculate this will calculate this K1 and will calculate this K2 and plug them back into this formula here and will find Y N+1.0247

So, and will step along will find T1 and Y 1 in T2 and Y 2 and keep going along by these little steps until you arrive at the value of T that you want to as that you want to use for your approximation of Y of T.0265

So, we will see how that works out it is a little complicated but it is it is it is not sort of theoretically difficult is just a matter sort of following these formulas through.0283

So, let us try that out in a first example really is run Dakota was step size H = .01 to estimate Y .01 in the initial value problem Y′ = 1+ T - Y and Y0 = 1 so, let me remind you of the Runge-Kutta formulas.0295

They run Dakota formula said TN + 1 = TN + H and Y N+1 = Y N+ H x K-1 + K2 /2 and K-1 was F T and YN and K2 that is the complicated 1 that is F of TN + H and Y N+ H K1 .0318

So, let us go ahead and try to figure those out here we have this the F is this function right here it is always Y′ = F of TY so, F of TY must be 1 + T - T - Y and we get our T01Y0 from the initial condition .0369

T0 is 0 and Y0 = 1 and so, now, let us start calculating T1 is T0 + H T0+0.1 which is 0.1 and out to find Y 1 we are to need to find these this K1 and K2 so, what is worked that out K1 is F of T0Y0 so, that is F of 0, 1 which is 1+0-1 which is 0 very convenient.0393

K2 is F TN + H so, that is F0.1 and YN + H x K1 so, YN was Y061 + H is 0.1 and K1 is all that is just 0 .0433

So, this is F of 0.1 and 1+0 so, just 1 and so, that is 1 + T - Y so, 1 +0.1 T -1 which is .1 so, that is our K-1 and K2 and now, we can find our Y1 is Y0 + H x K-1 + K2 /2 which is worth Y0Y0 was 1 + H0.1 .0458

Now, my K-1 was 0 my K2 was 0.1 /2 so, that is 0.1 / 2 is .05×.1 is .005 and so, I got 1 + that and so, he got 1.005.0510

So, my T1Y1 are 0.1 that comes from there and 1.005 and since I have already arrived at the value of T that I was looking for the stop and say that is my estimate for Y1 .005.0537

So, let me go back and recap here first I just copied the formulas from the beginning of the lecture that is what we found out at the beginning of the lecture for our new formulas, for TN + 1YN + 1 in terms of K-1 and K2 .0562

Now, the F comes from the differential equation that is F of T Y there my T0Y0, from the initial condition and my T1 is always just T0 + H so, that is 0.1 and then to find Y 1 I need my values of case so, unplugging everything into the appropriate F.0579

So, that is my K-1 that is my K2 playing everything into this formula for K2 that F is 0.11 and number this was my F so, it is 1 + T - Y1 +0.1-1 simplifies down to .1 so, now, I use my formula for Y 1 so, I get Y0 + H x the average of K-1 and K2 drop in what I figured out for my K2 and my K-1 was just 0 .0603

So, that is without 0 and out .1 come from work out there is a check I get Y 1 is 1.005 and since I already arrived at the value of T that I was asked for I got a Y value as an approximation.0636

So, example 2 works in any keep going with the previous example so, you definitely want to have watched example 1 before you try out this example is working to use our answer from example 1.0655

Reuse run Dakota was step size H = .01 0.1 to estimate Y .2 in the initial value problem is the same initial value problem we had for example 1 so, let me record first of all the formulas for Runge-Kutta our TN + 1 = TN + H our YN +1 = Y N+ H x this average K1 + K2 over 2 where the K-1is are given by K1 is just F of TN YN and K2 is more complicated F of TN + H T x T sub N + H and YN + H x case of 1 .0668

So, we are going to use our example from our answer from example 1 so, let me remind you what the answer from example 1 was if you have just watched it probably worth going back and looking at example 1.0725

Example 1 we figured out that T1 was 0.1 and our Y1 we figure this out before was 1.005 so, much as calculating those are coming from example 1 and so, let us go ahead and take it from there.0743

So, T 2 is T1 + H T1 0.1 H is 0.1 so, T2 0.2 and Y 2 our Y2 is before figure out Y to me go ahead and figure out what the case should be so, I will find K1 is F of TN Y Nthat is F of T1Y1 which is F of 0.1 and 1.005 .0758

Getting those numbers from example 1 and so, now, my F is this function right here 1 + T - Y so, 1 +0.1-1.005 and if I simplify that down it turns into 0.095 .095 so, now, I need to find my K2 its F of TN + H so, .1+ H is is .2 0.2 .0793

And now, Y N where was my Y N, that is 1.005 that is Y 1 + H0.1 x K-1 is 0.95 so, let us simplify that down in 2.095.0829

Sorry I put my decimal point the wrong place it is 0.95 .095 and so, if I multiply by .1 I get 2 0s .0095 and if we add that to 1.005 forget F of 0.2 and then 1.005 and .0095 so, 1.0145.0849

Now, my F function that is F right there so, it is 1 + the T value .2 - the Y value 1.0145 and so, that simplifies down .2-.0145 is .0 .1855 .0881

Sounds K2 and now, going to use my formula here to find Y2 Y2 is Y N Y 1 + H x K-1 + K2 /2 now, where is my Y1 there is 1.005 1.005+ my H is 0.1 K-1 was .095 and K2 is 0.1855 /2 .0910

And now, those are a bunch of decimals that I can just go ahead and throw into my calculator .095+ .1855÷2 x .1+1.005 and I see that when this all settles down I get 1.019025 .0950

And I can stop there the reason I can stop is because I have gotten to achieve value of 0.2 which is the goal T value that we are asked for the problem so, stop there and summarize T2Y2Y is T value is 0.2 and the Y value was 1.019025 .0985

So, that Y value right there is my estimate for Y of 0.2 and that is what we are asked for in this problem so, we remind you how all the steps worked out there.1011

I start out with the generic Runge-Kutta order to formulas also, known as approved Euler’s portals such as copies formulas overfill beginning of the lecture and those K1 and K2is are also, part of those formulas from the beginning of today's lecture .1029

This is the same initial value problem that we had for example 1 we did the first step of it in example 1 we found T1 and Y 1 so, example 2 here this can keep going with those values and so, I get my T2 is T1 + H which is 0.2 .1046

I am going to find K1 and K2 using my F function which I get from the differential equation so, there that is 1 + T - Y so, I plug-in T1 and Y 1 get .095 for K-1 for K2 I have apply my F function here .1068

I am reading off this formula right here and then plug in all the values TN + H there is my Y1 there is my H there is my K1 those simplify down 2.2 and 1.0145 and now, again using the F function appear so, I get 1 + T - Y simplifies down 2.1855 .1092

And then I played back into my main Runge-Kutta formula here sets run getting this line , and I just drop in all the numbers my Y1 my H and K1 and K2 worked out all those decimals on my calculator got 1.019025 and so, my T2 was 0.2 and that is my Y to write there and that is what I am stopping in the reason I am stopping there is because I have a Y .2 in the requirement for the problem.1118

So, since I party got into T = .2 then I take that Y value and I offer it as my approximation and as my final answer to that problem so, keep this problem in mind and also, keep this answer in mind because in the next example over to do is resolve this differential equation analytically and never go to compare our answer from the analytic solution with the estimate that we just made using the Runge-Kutta method.1153

So, let us see how those 2 answers compare so, example 3 we are to solve the initial value problem Y′ = 1 +2 - Y and Y0 = 1.1181

Resolve this initial value problem analytically in order to calculate how Y of .2 and really see how close the answer is to the estimate that we use in the previous example.1193

So, I cannot can have to do very much work in this problem because I have done this 1 before this is the same initial value problem that we did for example 2in the previous lecture on Euler’s methods.1208

So, if you do not remember that what you might want to do is go back and check out 2 to from the previous lecture on Euler’s methods so, from the Euler’s method lecture and we found the answer analytically there.1222

Remind you what was we found the answer was Y = C+E -T so, that is coming from the previous lecture not working that out on the spot here but if you have no idea where that is coming from just check out the previous lecture is example 2 on the Euler’s method lecture.1248

And so, now, we are going to compute Y of 0.25 0.2 = that if T0 .2+ E -0.2 and try that to signify that of course that is been recently used by calculator for 2130 So, 0.2+ E -.2 and I see I get 1.01873 that is approximation there so, that is my guess for Y0 .2 and that is really the answer given by an analytic solutions. 2203 They should think of that is essentially being an exact solution now, we also, solve this equation using the Runge-Kutta method and that was in example 2 of this lecture.1267

So, example 2 of this lecture so, that should be just above this 1 of this lecture we used the Runge-Kutta method to estimate Y of 0.2 and I am just a copy of the answer that we had back there got so, 1.019025 .1336

Sure that I am copying the right number there now, got 1.019025 using the Runge-Kutta estimation technique and we now, know, that the true solution is 1.0187.1370

So, let me just subtract those to show you how close we got there .0187 not cannot bother keeping track of decimal places after that because I see that is not 190 - hundred and 87 is 3 so, .0003 is our error there.1388

0.0003 is our error so, what you should really see here the new lesson you should take away from this is that the Runge-Kutta estimation technique is extremely accurate the true solution is 1.0187 our Runge-Kutta solution we figured out the previous example is 1.0190.1416

Very close the difference between those two answers is 0.0003 that is 1000 thousand 10,003 10,000 off from each other so, those answers are really very close so, even though it is a lot of work to calculate that Runge-Kutta solution the payoff is we get an answer that is very close to being the true solution is extremely accurate.1443

So, let us do a little more practice with calculating some Runge-Kutta answers in example 4 we are going to use Runge-Kutta was step size H = .0 .1 to estimate Y of 0.1 in the initial value problem Y′ = T-² + Y ² and Y001.1467

So, let me remind you of the Runge-Kutta formulas we have TN + 1 = TN + H and Y N+ 1 = Y N+ H + H x K1 + K2 over to this copying these back from the first slide of this as was the second slide of this lecture .1490

So, if you have not looked at the first couple size of this lecture that is where I am getting these formulas from where the K1 was F TN Y N and the K2 is as of TN + H and Y N+ H x K-1 .1525

So, let us apply those in this case Our F of TY that is just what you get from the differential equation so, that is T-² + Y ² and we start off at the initial condition that we are given so, that is Y 0s equal 1 so, Y01 start with T0, T0 = 0 and Y0 = 1 .1551

So, let me go ahead and plug all those numbers into the Runge-Kutta formulas so, TN +1 T1 is just T0 + H and that 0+0.1 so, that is 0.1 and now, I want to find Y 1 but in order to do that I need to know, my K1 and my K2 .1582

So, I got a fine K1 is F of TN Y N so, that is aft of T0Y001 and so, that is 0 ² +1 ² that is my F function 0 ² +1 ² is 1 my K2 is F of TN + H so, that is 0.10+.1 and Y N is 1 + H is .01 x K-1 is 1 so, that is 0.1 ² + now, 1 +.1 is 1.1 ² .1606

So, that is myself little more space here 0.01+1.21 so, that is 1.2 to select my K2 and out my Y1 is Y0 + H x K1 + K2 /2 now, my Y0 there it is this 1 by H is still 0.1 my K1 is 1 my K2 is 1.22 over 2 .1664

And if I just worked that out this 1 point this is 2.22 over 270.1×1.11 and so, that is .111 & get 1.111 .1710

And I can stop here because I got a T 1 of 0.1 I got a Y1 1.111 which is the T value is the T value those asked for the prompt for the problem so, stop there and summarize T1Y1 = 0.1 and 1.111 and so, that 1.111 that is my estimate for Y of 0.1.1729

So, that is what I would offer is my answer to the problem so, may I remind you where everything came from there for so, I just copy down these equations for the Runge-Kutta method to copy those down from the initial slide of the lecture 2nd slide lecture .1762

Then I identify what my F of T Y was that is just dysfunction that that we had for Y′ and then I identified T0 Y0 that came from the initial condition here and I started running through the equations.1778

T1 is T0 + H we are that is my H here so, that 0.1 my K1is F01 so, that is 0 ² +1 ² my K2 is F of this more complicated expression of getting that from up here by the way and I filled in all the numbers there so, that is T1 that is Y 0 that is H and that is K1 .1796

Filled in all those numbers get and plug it into my F function T² + Y ² simplify that down to 1.22 and finally use this YN +1 formula so, Y 0+ H x K1 + K2 over 2 there is Y 0 there is H there is K1 there is K2 and then I justify the decimals down to get 1.111.1825

So, my T1 and Y1 there is T1 came from their there is my Y1 and in a stop there and the reason I am been a stop is because I was asked to estimate Y of 0.1 and so, I am going to give my Y 1 as my estimate for Y of 0.1 1853

The end of this example we are going to use these this same initial value problem and the same computations as the first step of the next example 5 so, hang onto these numbers and can be the same initial value problem will take it 1 more step in example 5 .1876

So, in example 5 reuse Runge-Kutta was step size H = 0.1 to estimate Y in 0.2 in the initial value problem Y′ = T² + Y ² and Y 0s equal 1.1895

Now, once you remember this is the same initial value problem that we studied in example 4 so, we are going to use our answer from example 4 to solve this if you have not just watched example 4 just go back and check that out and you will see where we are getting our starting point for this for this example 5 .1914

So, let me remind you of our Runge-Kutta equations TN + 1 = TN + H YN +1 = Y N+ H x K1 + K2 /2and my K-1 = F of TN YN.1934

My K2 = TN + H and Y N+ H x K1 so, we are going to run this algorithm for this initial value problem but we party did the first step back in example 4.1966

So, go ahead and copy down the answers we had from example 4 might calculate those from scratch if you want to see where they came from just go back and watch example for you will see everything there.1988

So, what we had there is T1 we already figured it out back in example 4 T1 was 0.1 and our Y1 figure out last time was 1.111 .2000

So, really use those to take this out 1 more step so, our T2 is T1 + H so, that is 0.2 find my K1 K2 so, K1 is F of TN Y is that 0.1 and 1.111 now, my F function is given by the differential equation right here T² + Y ² that is F rate there .2015

And so, this is point .1 ² +1.111 ² and I do not think I want to do that plug that into my calculator so, .1 ² +1.111 ² is 1.244321 .2046

All say that as a number in my calculator K2 is F of TN + H so, 0.2 and now, my YN that is Y once a 1.111+ H is 0.1 x K-1 which is 1.244321 .2068

So, this is F of 0.2 and do some calculations right there so, looks like 1.235 here so, about 1.235 here and if I plug that into F I got .2 ² +1.235 ² .2097

And again I am going to calculate that out on my calculator and I see I get 1.566 for my K2 number ready to drop all those numbers and my formula for YN +1 find myself a Y2 is Y 1 + H x K1 + K2 over 2 and so, my Y 1 that is 1.111111.2129

H is still 0.1 and my K1 is 1.244 my K2 is 1.566 got all those numbers saved in my calculator now, and now, I am just going to simplify the decimals of them definitely do that on my calculator .2171

So, + 1.244321÷2 x .1+1.111 so, what I am getting on my calculator is 1.25153, in there some more decimal places after that but let me summarize here.2197

T2Y2 according to what we just figured out where is T2 right there, Y2 is right there so, T2 is 0.2 and Y2 is 1.25153.2231

And that is my answer as my estimate for Y of 0.2 and that is what we are being asked for the problem check this back out so, that is that means I am time of the problem below me go ahead and go back and make sure that all the steps are totally clear there.2249

We started out with the generic Runge-Kutta formulas copy those down from the second slide a lecture TN + 1YN +1 and K1 K2 complicated but not too bad and remember that this is the same differential equation we had for example 4.2272

We already did the first step of Runge-Kutta in example 4 so, that is what I am doing right here is just taking my answers from example 4 back watch example 4 and see where they came from and they we are to run with those answers and run 1 more step to get to get 2.2 .2289

So, T2 is just T1 +8 that is 0.2 and then I found K1 K2 using these formulas here K1 K2 and I plug in all the values that is T1 that is Y 1 that is T1 + H and that is Y 1 that was my H and this was K1.2311

And then to calculate F of those values running T² + Y ² so, that is Y doing .1 ² +1.111 ² and here I am doing.2 ² +1.235 ² .2340

So, calculate out those decimals and I get K1 K2 the drop those in the formula for Y 2 drop those in right there multiplied by Hs 0.1 and I still have my Y1 so, there is Y 1 there is H run using this formula right here to figure out Y 2 .2354

Drop in all those decimals and I justify those decimals of my calculator I did not think there is any reason to do that on the screen.2378

So, we get 1.25153 that is my T2 is 0.2 my Y2 is 1.25153 and so, since I have gotten to the value of T that I was asked to find in the in a prompt that are rather that I was asked to find my estimate at I am just to take the Y value at that point and offer that as my estimate of Y is 0.2.2387

So, my final estimate there is 1.25153 since the end of this lecture on Runge-Kutta order 2 techniques also, known as the improved Euler’s method if you seen it called that maybe in a class or in a different textbook.2418

That also, wraps up our chapter on numerical techniques when you cannot solve the differential equation analytically you try to use 1 of these numerical techniques Euler’s method or Runge-Kutta methods.2435

So, that wraps up this chapter and that puts another chapter of differential equations into the books so, I really appreciate your watching.2447

You have been watching the differential equations lecture here on educator.com. My name is Will Murray, thanks for joining us, bye bye.2457

Hi, welcome back to the differential equations lecture here on educator.com.0000

My name is Will Murray, and we are starting a chapter on partial differential equations.0004

So, the next few lectures are all going to be covering partial differential equations we will also get into Fourier series as well.0009

Before we really start partial differential equations I thought it would be worthwhile to review some partial derivatives just make sure that everybody is comfortable with the whole idea of partial derivatives so, that we can study partial differential equations.0016

Now partial derivatives are something that you really covering in detail in multivariable calculus so, if you are very rusty on partial derivatives what you might want to do is go back and look at the course on multivariable calculus on educator.com 0029

If you go back there it really cover partial derivatives in lots of detail, what I am doing this lecture is just going to give you a real quick crash course on 0103 Partial derivatives hopefully you have seen them before and hopefully will kind of remind you how partial derivatives we are.0046

So, let us go ahead and see what the definition of a partial derivative is the idea is that you have a function of 2 variables and sometimes this is given as F of XY for the sometimes these different variables are F XT I will use U of XT so, my function is going to be U my 2 variables are going the X and T.0067

If you are using different variables and your course of differential equations , then you just have to translate between X and T whatever variables you are using we are always getting used U of XT 0084

We will talk about the partial derivative with respect to X and the formal definition is in terms of a limit use these curly D here that is that is a curly D being on a regular old straight D like a regular derivative. 0096

We also, use U X to represent the partial derivative in fact that is the most common notation U of X because it is the simplest and the idea is that what we are going to do is walk over a little distance in the X direction.0111

So, that is why we change X to X + H and we are to see how much that changes the U values so, this is sort of like δ U / δ X so, be seen how much U change is will we change the X value by a little bit or just keeping the T values constant we are not changing the T values at all.0125

That is why it is the partial derivative with respect to X and the T so, we take that limit and that is the formal definition of partial derivatives.0149

In practice we will not really use that limit definition to actually calculate partial derivatives so, bit more in just a moment on how to calculate them .0157

Let me mention what they represent geometrically if you have a function U of XT and you graph it out in 3-dimensional space if you have X and T here then it represents a surface in 3-dimensional space .0166

So, I graph a surface in 3-dimensional space so, there is some surface and they that is the graph of U of XT and what the partial derivative represents geometrically is if you are standing on that surface and you walk in the X direction and represents the slope of the surface.0183

Walking strictly in the X direction so, holding the T variable constant and we are walking in the X direction the partial derivative the number that you calculate there represents that slope.0205

So, that is what it represents geometrically course that still does not really tell you very much about how to calculate it so, thought ahead and talk about how you calculate partial derivatives.0217

The way you do it is you treat the other variables of your finding U have actually treat the other variable T is a constant and then take the derivative with respect to X so, partial derivative with respect to X you just treat T is a constant and take the derivative just like you learning calculus 1.0227

What if you take the partial derivative with respect to T then you treat the X as a constant and take the derivative just like you did in calculus 1.0246

You can also, take second partial derivatives you can take U X X it means you take into partial derivatives with respect to X U XT would mean you take the partial with respect to X and then the partial with respect to T U TX is the other way around take the partial with respect to T and T the partial with respect to X .0255

UTC means you take 2 derivatives with respect to X with respect to T sorry .0275

Very nice property that says these 2 metal 1s the UXT and the UT X are the same there always equal to each other that is really nice what does it matter which order you take the partial derivative with respect to first.0281

If you are taking the mix partial derivative UXT does not really matter if you take the X derivative first and then the T derivative or if you take the T derivative first and then the X derivative .0300

So, you can do that in either order so, been doing this lesson is practice calculating a lot of partial derivatives and will also, check the theorem on a fairly complicated functions and make sure that it works out right.0313

You can also, use this as a check to make sure that your taking the derivatives right so, let us go ahead and see some examples.0326

In the first example it is a fairly easy 1 harder from here U of XT = X ² x T so, want to find the first partial derivatives U of X and U of T .0334

So, for U of X remember that means you treat T is a constant when we want to take U of X means you take treat T is a constant same thing in this X ² T is this being a constant x X ² .0345

So, the derivative of that is just that same constant x 2 X remember 2 X is the derivative of X ² with respect to X and the T comes along as a constant say to write that as 2 XT .0361

So, that is my U of X my U of T I treat the X as a constant which means X ² is a constant so, that just comes down when taken derivative because that is what happens with concerts and the derivative of T is just 1 so, the derivative with respect to T is just X ² .0376

Think of the X ² is being constant derivative with respect to T of T is just 1 so, that is my UX and my UT.0397

For that 1 just to recap there remember each time you take a partial derivative you are holding the other variable constant so, when you take U of X, that means T is constant think of T is being constant and so, that T comes down derivative of X ² is 2 X and you get 2 XT.0406

When you take the derivative with respect to T that means you think of the X is being constant so, X is constant constants and so, take the derivative with respect to T which means we have a constant x TX ² x the derivative is just that constant X ².0429

Let us go on and look at another example here that U of XT be sin of X x cosine T ² + 3T a little more complicated here .0454

To find the first partial derivatives U of X and UT so, remember we are going to find U of X first so, we can think of T is being constant which means cosine T ² is just 1 big constants.0469

Just and bring that down cosine T ² just 1 big constant now the derivative of sin X is okay derivative with respect to X is just cosine X now here is a mistake that lots of students make in multivariable calculus.0485

We have + 3T now remember if she is a constant that is just a constant so, the derivative of a constant is 0 so, that just drops out not 3T it is just 0 so, I write that is cosine X cosine T ² cosine X cosine T ² and I am done with finding the partial derivative with respect to X .0502

Now for U of T that means that I think of X is being constant so, sin X is just 1 big constant sin X but now I have the derivative of cosine T ² cosine is - side so, - sin of T ² - sin of T ² and now I have to multiply by the derivative of T ² so, that is 2T and I have + 3T and I am taking the derivative with respect to T so, +3 .0530

So, let me simplify this down a bit of the - 2T on the outside - 2T sin X x sin of T ² +3 and that is my partial derivative with respect to T .0566

So, recap how we found those things out when you find the derivative with respect to X your thinking of T is constant think of T is being constant so, that means that cosine T ² is nothing but 1 big constant .0588

Bring that down or take the derivative with respect to X a sin X we get cosine X and the key thing here is a T mistake that a lot of students make but I am training you not to is that the derivative of 3T with respect to X means you think of T is constant.0608

So, 3T is just a big constant and the derivative of a constant is 0 so, that drops out and just get cosine X x cosine T ² or take the derivative with respect to T that means you think of the X as being constant so, sin X is just 1 big constant.0627

That we have to take the derivative of cosine T ² remember cosine is - sin of T ² but by the chain role we have to put the derivative of 2T so, that is the chain role coming in right there and then the derivative of 3T is just 3.0645

So, if we sort things out we get - 2T times sin X + T ² +3 as our partial derivative with respect to T .0665

So, example 3U of XT is X ² + T ² will find all the first and second partial derivatives and we are in a check LaRose theorem in the context of this function .0675

So, when we find U of X start out with that used T is a constant so, I see through the X ² is 2 X derivative of T ² is just 0 because I think of T is a constant so, I just get just get 2X there .0690

U of T is the derivative with respect to T remember the X ² is a constant that goes to 0 and so, I just get 2 T now we take the second derivatives U X X is derivative of second derivative with respect to X so, I do the derivative of to X is just 2.0708

UXT is the derivative of 2 X with respect to T so, that is the derivative of 2X is just a big constant since worth checking the derivative with respect to T so, we just get 0 there.0733

Look over at UT UT X is a derivative of 2T with respect to X so, 2T is just a big constant now so, that 0 and UTT is the derivative of 2T with respect to T that is 2.0754

So, I found all my first and second partial derivatives the theorem says I want to check that UXT is the same as UT X and so, if I look at U XT and UT X they do agree with each other got 0 each time and so, hold at least for this function U .0777

So, let me recap there start out with X ² + T ² take the derivative with respect to X means the T is a constants would drops right out we get 2 X second derivative of that while the first rout of that is the second derivative of the original function is just 2.0801

But the derivative of U X with respect to T means you think of T is a constant so, the derivative of 2X is 0 on the other side UT think of X is a constants of the X ² drops out so, we get through the T ² is 2T.0818

Derivative of that with respect to X is 0 because the T is a constant derivative of 2T with respect to T is 2 check at UT X and UXT and making sure that we got the same thing both ways.0835

We get we got 0 either way so, UT UXT = UT X and example 4 we have a more complicated function U of XT = X / X + T again have to find all the first and second partial derivatives and check for this U so, I remind you here with me using the quotient rule a lot and if you don't remember the quotient rule we got a set of lectures on calculus 1 here on educator.com.0855

Go check them out and you will see lectures on the quotient role but meantime I have a too little mnemonic to remember the quotient rule.0889

If you think of the top as high / in the bottom is ho and then U′ it is the bottom derivative the top so, ho x the derivative of high - the top x through the bottom high x the derivative of ho may right be high x the derivative of ho / the bottom ² so, ho² and so, there is a cute way to say this.0897

So, that can be really useful for taking our partial derivatives Let us go ahead and try it out .0935

U of X is ho hi so, bottom are the top X + T derivative of X with respect to X remember everything here really thinking of X as our variable and T is our constant as long as we are differentiating with respect to X so, derivative X is 1 - hi is the top the derivative of the bottom is the derivative of X + T .0949

Since working derivative with respect to X that is just 1 and in the derivative of T is 0 all / ho of the bottom ² X + T ² and so, on the top of X + T - X that justifies down the T .0978

The bottom is X + T ² so, that is our derivative with respect to X .0996

The second derivative with respect to T see how that works out so, UT now many is the same quotient rule formula ho hi - hi ho / hoho but now I am thinking of T is the variable and X as a constant.1006

So, the bottom ho high X + T is at the bottom the high now the high part is X and the derivative of that with respect to T0 - high is X and the derivative the bottom is 0+1 all / the hoho of the bottom ² X + T ² .1024

So, this is X + T - X so, X + T is made some mistake here and I got a figure out what it is so, the bottom x the derivative of the top know I have not made any mistakes this is correct so, on the top and X + Tx 0.1050

That drops out - X / the bottom ² is X + T quantity ² that is my derivative with respect to T so, is my first 2 .1082

My 2 first partial derivatives and the problem also, says I need to find all the second partial derivatives So, find U X X you XT UT X and U TT .1097

So, let's work those out UX X you XX means I look at U of X and take its derivative with respect to X so, the bottom x the derivative the top ho-high X + T ² now the derivative the top is the top is T but I think of that as a constants that 0 - the top that is T x the derivative the bottom .1111

So, that is X + T ² derivative of that is 2 x X + T x the derivative of X + T using the power rule and the chain rule here.1136

So, derivative of X + T is 1+0 and on that now I have / hoho so, the bottom ² is X + T to the 4th and I see that this this term drops out because multiplied by 0 and now and X + T that will cancel with 1 of these and so, get a cube down there.1148

So, I see it simplifies down to is - 2T / X + TQ X + TQ so, that is my U of X X and now let me calculate U of X T so, ho high is the bottom x derivative of the top I am taking the derivative of U of X now.1174

Derivative the top is derivative of T which is 1 because T is our variable now X is our constant - the top x the derivative of the bottom so, the top x derivative of bottom is T x 2 x X + T using the power rule .1211

X + T ² is 2 x X + T x 1+0 and then / hoho so, X + T to the 4th looks little messy but I see I have X + T factor everywhere so, cancel out X + TX + T and 1 of my X + T is here.1229

So, I see what I have got here is X + T I see I had X + T on ² on the bottom so, this should been X + T ² the bottom x the derivative of the top .1251

So, if I cancel out the change that a little bit if I cancel out 1 X + T the whole thing does not cancel it just cancels out into X + T to the 1 so, let me fix that.1276

So, what I have got is X + T - 2T / X + T Q and so, X + T - 2T is X - T still / X + TQ.1289

So, that is my second mixed partial derivative U of XT we go to the other side and look at U of T and take a couple derivatives of that U of T X that means them to take the X derivative of U of T.1310

So, I am going to go ho-high bottom x derivative the top X + T ² is the bottom x the derivative of the top is the derivative of - X so, that is -1.1333

Since X is my variable right now - the top x the derivative of the bottom so, - - X x the derivative of the bottom is power rule okay 2 x X + T x the derivative of X + T with respect to X is just 1+0 all / the bottom ² all /is that was ho-high - high-ho X + T ².1350

I see you got X + T everywhere again so, we cancel out the X + T everywhere sorry the bottoms in the next should be x + T to the 4th.1381

Because we are squaring a ² and when I cancel out 1 of those X + T .1392

- X + T in the numerator - and - is + 2 X and in the denominator as you got X + T quantity Q and if I simplify that a bit on the top you got 2X - X lots of X - T all / X + T + not T there X + T quantity Q .1398

So, that is UT X something or U TT so, let me workout U TT so, you may go back to UT and take its T derivative so, ho-high x derivative of the top X + T ² x the derivative of the top and the top is - X.1434

But I am taking the derivative with respect to T now something of X is being a constant and is derivative is just 0 - the top x the derivative the bottom - - X.1463

So, now the derivative of the bottom is 2 x X + T x the inside derivative which is 0+1 all / hoho so, all / the bottom ² but the bottom by itself is X+t² when I ² a get to the 4th and so, that term drops out because multiplied by 0.1474

I see you got X + T here cancel of 1 of my X + T and I still got - X at with a - outside so, that is + 2X and I have got X + TQ on the bottom there.1505

So, that is my U of TT so, got all these partial derivatives and second partial derivatives now disposed to confirm Claro theorem says that U XT and UT X are supposed to come out equal. 1526

Let us compare those I see in each 1 of those boxes that I have X - T / X + TQ and so, it holds because those 2 boxes are equal U XT = UT X.1549

So, Claro's theorem holds those with a 2 boxes that we are supposed to be equal so, let me go back and show how we did each 1 of these.1567

First I wrote how the chamber of because are the quotient rule because we are using the quotient rule over and over again .1589

My version the quotient rule is ho-high - Heidi Ho / Ho ho that is shorthand to help you remember bottom x the derivative of the top - the top x the derivative the bottom all / the bottom ² .1608

So, we apply that to the initial U when we take derivative with respect to X for that means T is constant T is constant and so, we work out the bottom x derivative of the top X are variables of the derivative is is 1 top derivative the bottom and the bottom ² and that is if I down the T / X+T² .1610

That we take the derivative with respect to accept that so, again T is constant and so, we get the bottom x the derivative of the top but since the top was T and T is constant that is what we get that 0 from .1633

And then to find the top x the derivative of the bottom that is the top right there T remember the bottom we have to use the power rule so, that is as the derivative of that anything ² is 2 x that original thing x the derivative of that things we are using the chain rule as well there.1650

And the bottom ² a Ho Ho is X + T to the 4th but then that X + T and that X + T cancel each other out so, that is why we ended up with X + TQ in the denominator in our final answer.1672

We did UXT we start with UX we started with T / X + T ² but then we held X constants and we took the T derivative X is constant and ran through the quotient rule again so, there is my Ho D high and high and all of this is the Ho so, ho-high - Heidi Ho and again we are holding X constant and taking derivatives with respect to T and there is my Ho Ho right there.1688

But when I look at this I can cancel out and X + T from everything and so, that simplifies down to just a single is X+T² to single X + T - 2T / X + TQ and then that simplifies a little more than numerator simplifies X + T - 2T just reduces down to X - T / X + TQ.1726

That was UXT / on the other side finding U of T, that means X is constant so, we do ho hi the D high gives me just 0 there because the derivative of X is 01 we are assuming X is constant - Hi D Ho so, that is that is the high part and that is the D Ho part.1748

And then that is Ho Ho right there since and find down to - X / X + T ² and we take the derivative of that with respect to X settings we are holding T constant T is constant.1779

So, we take the derivative of that using the quotient rule again so, there is our Ho the high since our X are variable now derivative of X of - X -1 - the high and there is her de-ho again using the chain the chain rule and the power rule.1791

And there is Ho Ho but again we have an X + T canceling everywhere so, we go down to X+T cubed in the denominator numerator simplifies to - X + T just a single power there cancel off 1 power and + 2X in the simple find down to X - T / X + TQ only take the T derivative of UT holding X constant.1814

And so, we are again getting is the chain rule but what we do so, Ho x D high D high .1840

High was X but since X is constant derivative is 0 - there is my high and there is my de-ho again so, Heidi Ho and then there is Ho Ho so, hody high - Heidi Ho / Ho Ho.1856

So, Ho Ho is X + T the 4th but again we have X + T canceling so, just goes down to X + TQ wind up with to X / X + TQ .1871

That is we are all those partial derivatives came from the last step in this problem was to confirm Claro theorem which means your checking to see whether UXT is the same as UT X.1881

Now not all these derivatives we are the same but if you look at U XT here and UT X we did calculate those independently but after we some find them down we got the same thing on both sides.1892

Got X - T / X + TQ and so, we can say for sure that Claro theorem did hold for this particular function.1906

In our example 5 we are going the opposite direction from taking partial derivatives this time we are given a partial derivative or asked to figure out what the original function could be that have used that.1918

So, essentially what we are going to do is working integrate with respect to X so, I am to integrate = the integral to think of it this way as EXT cosine T the X at but now that means we are integrating with respect to X.1929

So, T is constant and just like when we took the derivative with respect to X the help T is constant and so, that means that cosine T is just in the constant which means you can pull it out of the integral.1950

So, cosine T is not equals and pull that out of the integral E XT DX now really integrating that with respect to X me just remind you for example E to the to the 4 X D X would be 1 4th E 4 X + the constant that is not going back to the old calculus 1 technique of substitutions of that looks foreign to you.1963

Check back to calculus 1 lecture here on educator.com it is doing a little substitution U equals 4 X and DU = 4 DX and then doing that little substitution and up with 1 4th in the 4 X.1995

So, here we got cosine T now you want to think about this as E the T X where T is a constants was kind behaving like the 1s the 4 here so, is that 1 4th E 4 X .2032

I get 1 / T x E TX just like the 4 was before now we have T + now be careful here I want to say + C but remember I am integrating with respect to X which means that I am thinking of all T as being constants.2039

So, what I am going to say is + any function of T because if I took the derivative of that if I took the derivative if I took U of X of that it would just go back to 0 so, can really have any function of T here that I like .2049

So, let me collect my terms and simplify that 1 / T x E TX x cosine T + any function of T so, this could be any function of T this is my U of XT any function of T could be your CT here and the reason I can put any function of T in their cosine T in the T natural on the TT ² is because if I took the X derivative it would just cancel away to 0.2066

So, that is my most general form for my U of XT if he took the derivative the X derivative you get back to what we started with .2108

You might ask where is the constant here you can think of the constant is being built into the function of T so, this includes that any possible constant that you would care to add on there.2123

So, let me recap how we figure that out.2140

Basically we are doing the integral but we are doing the integral with respect to X and so, that means that I can think of cosine T as being a constant and I can plug pull it out of the integral and I am integrating E as even the TX remember T is a constant so, the integral E TX is just 1 / T x D TX .2143

Still have that cosine T and normally I would tack on an arbitrary constant here but since I am doing up the opposite of a partial derivative I contact on any function of T because any function of T be treated as a constant.2164

When we take the partial derivative so, this would be considered would be treated as a constants when taking the derivative with respect to X .2181

When finding U of X so, any function of T could be included there and so, just can I include that arbitrary function of T C of T instead of including an arbitrary constant and I can think of an arbitrary constant being built into it.2204

So, my final answer there is what I got from the integral + any arbitrary function of T and the reason that works is because if you took the X derivative that C T would just completely drop out and we get back to that in the E XT cosine T that we started with .2221

So, that wraps up our review of partial derivatives will be all set to go now for learning about partial differential equations starting in the next few lectures.2239

If this was not enough of you partial derivatives, if you are still feeling a little rocky little bit rusty and when you taking partial derivatives then we have a whole set of lectures on multivariable calculus and you get more practice from taking partial derivatives in those lectures.2252

You can go back and watch those lectures again the 1s on multivariable calculus you get lots of practice with partial derivatives.2268

This was just meant to be kind of a quick review practice brushing the rust off so, that when we start doing partial differential equations in the next lecture you will be ready to go.2275

So, that is that is the end of our lecture on reviewing partial derivatives and you are watching the differential equations lecture series here on educator.com.2286

My name is Will Murray, and I very much appreciate your watching, take care.2297

Hi and welcome back to the differential equations lecture here on educator.com. 0000

My name is Will Murray , and today we are studying partial differential equations.0005

Meet some of the most common partial differential equations , today focus on the heat equation we are to look at boundary value problems in some of the associated conditions are often given as part of a package with differential with partial differential equations.0010

To learn how to interpret each 1 of those conditions in terms of physical setting so, let us get started.0023

So remember, a partial differential equation is an equation that relates a function of 2 variables which were to call U of X and T X and T are the variables and U is the function.0030

So, partial differential equation relates U with 1 or more of its partial derivatives.0043

So, we did a whole lecture reviewing partial derivatives so, I hope U are comfortable with partial derivatives by now, if you are not comfortable with partial derivatives maybe go back and watch the previous lecture here E differential equations lecture series.0049

Got a long review of partial derivatives, if that lecture is not enough then you might want to go back and watch the videos from the multivariable calculus where they really do partial derivatives in more detail.0063

So, the most common partial differential equations have the second partial derivative of U with respect to X, the U of X-X and then I gets related to either the partial derivative with respect to T or the second partial derivative with respect to T .0075

So, we will meet a couple of those are common partial differential equations in just a moment let me tell you about at boundary value problem .0092

A boundary value problem is a partial differential equation which comes together in a package with initial conditions and boundary conditions.0099

Often, the boundary conditions or are 0 so, we show let me show you what I mean by that. 0109

The initial condition is this first condition here the idea there is that the plug-in time = 0 that is why it is called the initial condition is because se it takes place at time 0.0114

So, we plugged in T = 0 and so, we just get a function of X so, this would be given to you all these context of the problem all of these equations would be given to you.0130

The boundary conditions tell you what happens when X = 0, when X = some Extreme value which were usually going to call L so, we plugged in X = 0 or X = L which is a constant.0145

And the boundary conditions tells you what happens at different times at the edges of your interval at 0 on one side and it fell on the other.0160

So, these collectively are boundary conditions and will see how those get interpreted physically but the important thing to know, right now, is that L is a constant and all of these of this initial condition and these boundary conditions these would all be given to you any package along with the partial differential equation.0171

The goal is to find a function U of XT satisfies all of those.0198

We will see how to how to solve that over the course of the next few lectures in the meantime I just want to study these and this initial condition these boundary conditions little more carefully and see how we interpret them physically.0205

So, let us introduce the most common partial differential equations.0218

For 3 very common 1s are known as the heat equation and the wave equation and Laplace’s equation each 1 takes a quite a long time to really study and solve.0224

So, you kind of an studying the same equations over and over again once you learn each 1 then you really have a good grip of partial differential equations.0234

So, the heat equation says U of T remember that is the partial derivative of U with respect to T = α ² x U of XX.0243

What that α ² is the constant and I will say what that measures in just a moment U XX is the second partial derivative of U with respect to X and what this equation describes is a function U of XT which tells you the temperature of a solid rod of rod could be of different of different materials it could be a metal rod it could be a wooden rod we often think of it as being a metal rod.0253

And it tells you the temperature of a rod at position X at time T and X here is the position along the rod E left hand of the rod is being X = 0 the right-hander the rod is X = L E rod is L units long and so, were trying to solve this discover what happens for all positions X between 0 and how and for all time T.0288

And the ideas usually that you know, the initial distribution of heating the rod and then gradually the heat diffuses throughout the rod in a cool down and you are trying to find a description of what temperature each place along the rod is that each time.0318

So, that is the heat equation very common partial differential equation that is the 1 were actually going to study and solved in the next few lectures here.0334

The wave equation is another very common 1 and it looks very similar U TT = C ² x U XX and again C is a constant.0343

I forgot to mention back were talking about the heat equation that α ² is a constant and it is a constant of thermal diffusivity which means it is a constant that depends on the material in the rod.0354

If the material in the rod conducts heat very easily than you have a higher value for α ².0371

If the material in the rod conducts heat poorly then U have a lower value for α ².0377

So, this let me interpret the wave equation for you the idea of the wave equation is you have a string that is attached at 2 ends so, again we have X going from 0 to L and this string what you do is you pluck it and so, it vibrates back and forth .0388

So, the string vibrates back and forth as time goes by and so, what you have is the string has a particular position particular amount of displacement at any given time.0406

So, this displacement depends on where you are on the string and on what time it is and then U of XT describes the displacement of a string from 0 at time T.0428

Now, that C ² that constant is a factor of how elastic the string is of course if the string were so, completely rigid object if it were just a metal rod then it would not vibrated all.0439

If the string were something very very elastic like a rubber band for example it is going to vibrate a lot and so, that C ² is a measure of how elastic the material in your string is.0455

The last equation that is very common when you study partial differential equations is Laplace's equation which says that U XX + UTT = 0 and this 1 comes up a lot in complex analysis.0470

I am not going to interpret that 1 physically for you because se it is not 1 that were going to be studying later we are to spend most of our time studying the equation later.0483

I just want to show you Laplace's equation because these really are sort of the big 3 partial differential equations that people study the heat equation the wave equation and Laplace's equation.0492

So, let us see how the initial conditions and boundary conditions affect the physical situation here.0504

So, let us jump in some examples the first example this is give us a physical interpretation for each element of the boundary value problem associated with the heat equation than what we have here is a partial differential equation and then several boundary and initial conditions.0513

So, let me show you how to interpret each of these each of these equations here I told her that α² is a constant that depends on how easily the material rod conducts heat so, it depends on the thermal diffusivity of the rod diffusivity of the material in the rod.0529

Just to give you an idea if the material is very conducted then you get a higher value there for example Silver has α ² = about 1.7 whereas if it is lower if it is a material with lower conductivity than you have a lower value of α iron for example does not conduct heat as readily as silver does.0579

So, iron has α ² = 0.12 so, you get different values of α depending on depending on how conducted the material in Ur rod is.0612

Now, let us look at these other conditions the initial conditions in boundary conditions let me draw rod here and you want to think of X going from 0 left hand side to L on the right-hand side now, U of X-0 that means we plugged in 0 for T time 0 = FX .0626

So, what that really tells you is the distribution of heat in rod when we begin experiment so, that is something that an experimenter data a physicist or scientist would have to go and measure physically before starting the experiment.0651

So, that is the initial temperature of the rod of U of 0T that means we plugged in X = 0 which means were looking at the left end of the rod here we were being told is a U 00T = 0 for all time.0670

So, that means the left end of the rod is being kept at temperature 0 for all time so, you want to think of that as the left end is being To temperature 0 so, it is being artificially forced to stay temperature 0 perhaps by being packed in ice.0697

So, no matter how much heat might travel to that left rod say from the middle of the rod were forcing the left end of the rod to stay at temperature 0.0718

U of LT that means we plugged in X = L which corresponds to the right end of the rod if that is also, forced to stay at 0 that would mean that the right end of the rod is also, packed in ice.0730

So, all of these conditions would have to be given to you at the beginning as part of a package which is known as a boundary value problem given all of these conditions then you could start solving at solving the partial differential equation trying to match the initial conditions the boundary value conditions and we have not learned how to do that.0745

That is going to be the content of the next several lectures here on educator.com so, this point were just kind of learning how to interpret each 1 of these conditions what they need physically and learn how to solve the next time.0773

So, let me recap there what we said was this UT = α ² UX X the partial differential equation and the only thing that ever changes there is the α ² and even that never changes very much because se that is always a constant it just depends on the material of that your rod is made out of the U of X-0 that means U plugged in T = 0 so, that means you are looking at the initial temperature of the rod .0784

Again this would all be given to YOU so, FX would be a function that the experimenter has determined by going through in measuring the initial temperature of the rod and then U of 0T and U of LT that means U plugged in X = 0 and X = out which means you are looking at the left-hand end of the rod.0831

E right-hand end of the rod and if for keeping them = to 0 that would suggest that leaves the artificially forced both ends to stay at 0 perhaps by submerging both ends of the rod and ice bath and so, that is why how we know, the temperature 0 for all time .0840

So, let us look at a different kind of the boundary value problem still heat equation in example 2 were still looking at a heat equation but we have been given slightly different conditions here so, let us see how to interpret your physical interpretation for each element of this boundary value problem .0855

So, we have U of T = 1 4th U of X X that is the same heat equation is before Except we been given a specific value of α ² here and that is a constant which measures how susceptible the material E rod is to conducting heat.0872

Constant of thermal diffusivity and I switched these around the little that the U of X-0 means we plugged in T = 0 so, that is the initial distribution of heat in rod or the initial temperature in rod of the rod distribution of heat in rod .0895

So, this is something that an Experimenter would have to go through and measure the temperature at each place along the rod and tell you the mathematician what the temperature is at each place at each position X and at each time well at time 0 .0931

Then, here is the new element for this problem is set of U sub0 is that a 0T we have U of X-0T so, how is that different well what that means is that if U of X of 0T means we plugged in X = 0 so, looking at the left end of the rod and if U of X = 0 that means the rate of change in X direction is 0.0957

Which means there is no change in X direction so, this is telling us that there is no change in X direction so, there is no heat flow this is at the left end of the rod because se X = 0 no heat flow across the left end of the rod.0991

Which means we somehow prevented any heat from flowing in or out of the left end of the rod which means we must a packet in some kind of insulating material.1023

So, the left end is packed in insulation now, I want to emphasize the distinction there before we had U of 0T was = to 0 which with the temperature was forced to be = to 0 now, we had U of X of 0T = 0 that means the temperature flow the rate of change E temperature is forced to be 0.1038

So, this is really affect the U of X being = to 0 there is no change in temperature across the left end of the rod before we packed the left end in ice there could be a lot of heat sort of flowing out into the ice bath here we are saying that there is no heat flowing in or out so, we must be packing in some kind of insulating material to prevent any change in temperature across the left end.1068

Now, we plugged in X = 1 for the second condition these are both boundary conditions because se they tell us what is happening at the edges of the rod boundary conditions and so, since we plugged in X = 1 that must mean L = 1 must be dealing with a rod of length 1 and since L = 0 that means there is no heat flow across the right ends of the rod .1095

So, that right end must also, be packed in insulation so, what were doing here is we are actually solving the problem or not doing any math at this point were just learning how to interpret each of these elements of a boundary value problem that would all be given to us as part of actually solving a heat equation .1131

So, the beginning here this was the in this is always the same except that that value of α might change will always be constant and it is a constant reflecting the kind of material you have in your rod.1169

Material that conducts heat more easily would have a higher value of α ² material that is less conductive of heat would have a lower value of α ² .1188

Now, the difference from this problem to the previous 1 from example 1 is E previous 1 that was adjusting U in here we have a U of X what that represents is that at the left end of the rod X = 0 and the right E rod X = 1 we got no change in temperature across the boundary.1196

Which means that there is no heat flowing in or out across the boundary which must mean that we have artificially preventing heat from flowing in or out which means we must have packed it in some kind of insulating material.1217

That is how we ended up concluding from these boundary conditions of the left end and the right end must be packed in insulation .1230

Finally we have this initial condition here which says that U of X of X, 0 = some function of X that is telling you the initial distribution of heat throughout the rod so, that would be something that the Experimenter has to go and measure before presenting all of these conditions together as a package called a boundary value problem. 2100 So, example 3 were also, going to give a physical interpretation for each element of the boundary value problem associated with the wave equation so, let us see how to interpret each 1 of these elements of the wave equation.1238

U TT = C ² U X X that is just the partial differential equation for the wave equation and the only thing that ever changes there is that C ² is always a constant but it might be different constant depending on what kind of material you have in your string.1276

So, remember the wave equation measures the displacement of a vibrating string at time T and it position X so, X but varies between 0 and L and T goes from 0 to infinity because se we want to watch this string sort of oscillate infinitely.1298

Now, that C ² is the constant of elasticity it tells you how elastic the material of your string is in other words if if you are dealing with for example when the rod instead of a string that is not going to be elastic at all that is not going to vibrate up and down at all.1333

If you are dealing with something more flexible like a rubber band that something is going to be very elastic and so, would have a much higher constant of elasticity. 2248 Now, U of X-0 that means we plugged in T = 0 so, that means were looking at an initial condition here in initial condition and so, that would measure the initial displacement of the string .1360

Maybe if you taken the string and you stretched it out beautifully plucked it and you are about to let it go that F of X would tell you the position of the string at any given point at time = 0 .1388

That U of X-0 as U of 0T means we plugged in X = 0 which means were looking at the left end of the string and if that= 0 that means it is not displaced at all for all time which means you pin down the left hand and that the string.1401

So, left end you pin down the left end of the string of string is pinned down and then the same kind of thing here we plugged in X = L into U of LT and we been told that U LT = 0 so, that means the right hand and of the string is also, pinned down for all time.1419

So, what all these equations mean the first 1 is just the PDE which tells you the general behaviour of a vibrating string and the C ² is a constant of elasticity which depends on the material of the string .1455

Is it actually string is it guitar wire is its rubber band is it something completely rigid like oh a wooden bar so, you get different constants of elasticity limit with the context of any 1 problem that number will always be constant so, it will stay fixed it C ².1482

Now, the other conditions here are essentially keeping track of everything U would need to know, about in order to do is determine and predict the behaviour of that string U of X-0 tells U the position of the string when you initially pluck it or when you measure it at time 0.1508

Such a stretch it up and then let it go when U stretch up measure the initial position of that string and that would B F of X that so, that is the initial condition U 0T and U of LT means U plugged in X = 0 and X = L and what it is telling you since their 0 is that your keeping the left hand and E right hand then pinned down for all time .1526

So, let us go ahead and look at some possible solutions to differential equations so, were to look at the solution to a possible solutions to the wave equation U TT = 9 U X X .1551

So, what were going to do here is look at each 1 of several different types of functions and I may calculate U TT for each 1 of them may calculate U X X , for each 1 and were to see if they work out this post workout being told to check that each 1 works out.1568

So, we will see for each 1 of these functions the first 1 is U of XT = cosine X sin of 3T so, let us figure out U of TT and then we will figure out U of X X for each 1 of these functions.1589

Now, cosine X x sin of 3T we figure out U of TT that means that this cosine of X want to think of that is being constant and so, we will just have cosine of X coming down now, sin of 3T were going to take 2 derivatives of that so, the first derivative of sin is cosine a 3T and then we multiply on a 3 by the chain rule.1608

That is the first relative and E second derivative of the negative sin of 3T and we multiply out another 3 so, we end up getting -9 there and so, U of TT would be -9 cosine X x sin of 3T .1639

Now, U of XX works just the other way around now, we think of sin of 3T is a constant so, sin of 3T is just going to come down the constant but cosine X me to take the second derivative of that while the first revenue cosine X is negative sin X the second derivative of cosine X is negative cosine X.1664

So, U of X X is negative cosine X x sin of 3T and if you compare these to each other what we see is the U of TT is exactly 9 x U of XX and so, it satisfies the differential equation so, we just checked that our first function there satisfies the partial differ differential equation.1687

So, it is going try that with the second function here U X at U so, that is T = cosine X U of cosine a 3T so, U of TT is cosine X because se we think of that is just being a constant for taking the derivative with respect to T cosine a 3T only take 2 derivatives of that .1710

The first relative is negative sin of 3T multiplied by 3 by the general and the second derivative through the sin is cosine a 3T but we saw that negative and now, we get another 3 popping out so, -9 cosine 3T so, -9 cosine of X cosine of 3T down the other side we were looking at U of X X so, that cosine a 3T would just come down as a constant since you think of T is being a constant .1735

Now, through a cosine on the second cosine this negative sin derivative of that is negative cosine her second derivative is negative cosine X x cosine a 3T again if you compare those 2 to each other on the left we have negative 9 x what we have on the right.1769

So, U of TT is indeed = to 9 x U of XX let us try that again with the third function here U XT = E T x E X over 3 U of TT means we think of X is a constant we take the second derivative of ET which is just E T and then E X over 3 is just comes down as a constant .1790

Each of the X over 3 U of X X means the E T comes down the constant or take the second derivative of E X over 3 will the first relative is E X over 3×1/3 .1821

Then the second derivative is E X over 3 multiplied by another thirds we get 1/9 so, 1/9 U of T x E X over 3 and again we can see that the left-hand side = 9 x the right-hand side so, if we compare those to each other they definitely do satisfy the partial differential equation E T x E X over 3 is 9×1/9 E T x E X over 3.1837

Try again with U of XT U of XT = E 3T x E X so, U of TT means the X can be a constant E 3T second derivative of that, first derivative is 3 E 3T second related 9 E 3T.1866

So, we get 9 E 3T x E X over on U of X X that means E 3T is a constant and the second derivative of E X is just E X and so, I wrote a 3 here it should have be 9.1887

So, what we see comparing the left-hand side the right-hand side the left-hand side really is 9 x as big as the right-hand side so, looking back up our differential equation U of TT = 9 U of XX.1913

Finally in our last 1 we have U of XT is = 3T + X we calculate that 1 a little more carefully U of T = the derivative sin x cosine of 3T + X x 3 the derivative of the inside and then U of TT is a derivative of cosine is negative sin of 3T + X .1928

So, 3 x that first 3 would be 9 now, let us look at U of X U of X is cosine of 3T + X and U of X X is just negative sin of 3T + X.1955

You do not have to worry about the chain rule here because the derivative of the inside stuff with respect to X is just 1 .1982

My U of TT is -9 sin of 3T + X is copying that over from left there my U of X X is negative sin of 3T + X now, if you compare these 2 to each other you see again that the left-hand side is 9 x as big as the right-hand side and so, again we have a solution to the partial differential equation.1989

So, in each case we have U of TT = 9 x U of X X so, all of these functions work as solutions to the wave equation U TT = 9 U of XX .2020

So, let me remind you how we did that in each 1 of these functions we worked out U of T and U of TT and U of X and U of X X and then I just kind of listed what we got on each of those for each of those second partial derivatives over on the right here.2034

You are pointing to remember when you are taking this these partial derivatives is that whatever variable you are differentiating with respect to you want to hold the other variable constant.2054

So, for example when I took the derivative with respect to T I hold the X part constant so, if even if I have cosine a 3X or something like that it just comes down as 1 big constant so, what I did was take the second derivative with respect to T secondary with respect to X I did that for each 1 of these functions holding the other variable constant each time .2064

I checked on each 1 that what I got on the left for U of TT really was 9 x when I got on the right for U of XX and it worked out in every1 and so, I can say that each 1 of these functions really is a solution to that wave equation .2089

Example 5 where the check for each 1 of the following functions as a solution to the heat equation U T is = 4 U X X were going to also, determine which 1 also, satisfies the boundary condition U of 0T = 0 .2114

So, let me first just work out UT and U X X for each 1 of these functions so, let us look at U of T and U of X X for each 1 of these functions so, for the first 1 U T would be E X of a constant x E 4T x the derivative of 4T so, that would be E X E 4T U of X X just means you are taking the second partial derivative with respect to X which the second partial derivative of E X is just E X and then either 4T comes down as a constant .2129

And if you compare both sides here it does look like the left-hand side is 4 x as big as the right-hand side so, we do have a solution to our heat equation therefore partial differential equation.2175

In our second function here got E X over 2 on the left and E T on the right so, U of T is just the derivative of E T because se E X over to as a constant so, E T derivative is just itself some nothing really happened there.2185

U of X X well take 1 derivative here we get 1 half E X over 2 if you take a second derivative 1 half x 1 half so, 1 4th E X over 2 then E T just comes down as a constant so, we get 1 4th E X over 2x E T .2204

If you compare those to each other you do see that the left-hand side is 4 x as big as the right-hand side so, again we have a solution for partial differential equation for part C if we do U of T well looks like the sin X comes down as a big constant and the derivative of E negative 4T is -4 E negative 4T .2228

Then if we take the second derivative with respect to X the E the 4 negative 4T just can be a constant but sin X the first derivative is cosine X the second derivative is negative sin X derivative of cosine is negative sin X and so, we get negative sin X E negative 4T .2256

Now, we notice if we look at the bulk that both sides there the left-hand side is exactly 4 x the right-hand side and so, again our U of T is 4 x U of X X so, we do have a solution to the partial differential equation.2283

Let us get look at U of XT = to cosine of X over 2 E negative T so, our U of T just has cosine of X over to coming down as a constant E negative T the derivative of that is negative E negative TU of X X the E negative T is going to come down as a constant the derivative of cosine X over 2 is negative sin of X over 2.2301

Then we have to multiply by 1 half by the chain rule and derivative of sin is cosine of X over 2 multiply by 1 half again by the generals we get -1/4 get -1/4 cosine of X over 2 x E negative T and again if you compare the left-hand side of the right-hand side each the left-hand side is exactly 4 x the right-hand side .2329

So, again we have a solution to the partial differential equation there so, let me recap what happened here each 1 of these functions will we did was we tried it out in partial differential equation when we tried out was just by calculating the T derivative holding X constant and then the second X derivative holding T constant .2365

So, here all the derivatives that we calculated on each side each time and then for each 1 we checked that the T derivative the U of T was exactly 4 x the U of X X and it did not work out every time and so, each 1 of those really was a solution to the partial differential equation that we were given .2391

Now, our X also, to determine which 1 also, satisfies the boundary condition U 0T = 0 so, we do that on the next slide so, next slide is where we’re going to try to satisfy the boundary condition U 0T = 0 .2411

Let me remind you of the 4 functions that we had, here are first 1 was U of XT = E X E 4T our second 1 was U of XT = E the X over 2 x E T third 1 is U of XT = sin X x E negative 4T and our last 1 was U of XT is cosine of X over 2 x E negative T now, for each 1 of those were in a plug-in X = 0.2433

So, let us try plugging in a single 0 for each 1 of those who would get E 0 x E 4T which is just in 0s once was E 4T here again we get in 0 x E T which is just E T here we get sin of 0 x E negative 4T.2484

Now, sin 0 0 so, this just cancels a way to 0 and finally here we get cosine of 0 x E negative T now, cosine of 0 is 1 so, this is even negative T and what were looking for is 1 where U 0T comes out 30 so, clearly the first 1 fails the second 1 fails the third 1 works and the 4th 1 fails .2511

So, that is our solution right there that is the 1 that satisfies both the partial differential equation and the boundary condition X 0T = 0 so, just to remind you what we did for each function there.2536

We are plenty of X = 0 so, each 1 of these were finding U of 0T plug-in X = 0 and simplifying it down U 0T U is 0T U are starting to look like Y for me I clean up those U there .2554

For 1 of those the function simplify down to be = to 0 which is what we are looking for and so, were to say that that 1 is the actual solution to both the differential equation and to the boundary condition at U 0T = 0 .2579

So, that wraps of this lecture on the partial differential equations and in particular the heat equation we remind you that we did not actually learn how to solve the heat equation in this lecture we will learn that in the next lecture .2596

We relearned about Fourier series so, and actually that I think it is 2 lectures for now, word to learn about Fourier series next lectures can be on separation of variables.2610

So, let me just remind what we did in this lecture will learn how interpret all the different conditions that come package with a partial differential equation.2622

We learn what each 1 of them means physically been actually learn how to solve a partial differential equation but I hope now, that you are little more comfortable with interpreting each 1 of those initial conditions and boundary conditions that you would receive as part of a boundary value problem along with a partial differential equation.2630

So, hopefully you can interpret each 1 of those you know, what each 1 of them means and in the next few lectures were going to learn about separation of variables in order to learn about Fourier series.2650

We will actually learn how to solve all that together and solve the heat equation.2662

So, in the meantime you have been watching educator.com , these are the differential equations lectures and my name is Will Murray. We will look forward to talking to you in the next lectures. In the meantime, thanks for watching, bye bye.2667

Welcome back to the differential equations lecture here on educator.com.0000

My name is will Murray, we are studying partial differential equations and were starting to learn how to solve them we already had a couple lectures sort of warm you up to the idea of partial differential equations.0003

We have not actually solved that yet so, today were going to learn how to start to solve them using a technique known as separation of variables so, see what that is all about.0016

The separation of variables like as I said , it is a technique for solving some partial differential equations and the idea is that you assume that the function you are looking for remember work in a call that function U of X of T can be written as a product of a function of X only and a function of T only.0027

So, we assume that the function were looking for can be written as were to call it X of X of X and T of T by the way were to have lots of equations today where were using X is a little x and T little t .0048

Be very careful to keep those straight the little X and little T are the variables X and T are the functions so, let me emphasize that right now, those are functions and the little X and little T are the variables try to be clear when I am writing my own notes which 1 is I am talking about.0066

And so, the point of assuming that you have a product of 2 functions like that is it is really easy to take derivatives because of your taking the partial derivative with respect to X that means that all the Ts are constant so, T of T just comes down as a constant we think of that as being a constant .0091

So, you just take the derivative of capital X with respect to little X X′ of X and then when you take the second derivative with respect to X your taking capital X″ of X and again the T is just a constant when you are looking at derivatives with respect to T is just the other way around .0111

The capital X you think of as being a constant and your taking derivatives of T with respect to little T so, here is the first derivative is T′ and in the second derivative is to″ .0134

So, you take these derivatives and you plug them into the partial differential equation and then let us see what happens with that what you trying to do is get all the X is on 1 side and all the Ts on the other.0149

It does not always work but it is it works for a lot of these differential equations will see some example lots of examples where does if you can do that then a very interesting thing happens you have 1 function that is only dependent on X 1 function that is only dependent on T and are equal to each other.0162

That means if you change T nothing happens on the left because X has not changed so, if you change T nothing happens which means it must be constant so, both of these functions must be constant so, and I call that constant .0184

And so, were to end up with a function of X = λ and independently a function of T = λ so, will reorganize these into 2 ordinary different differential equations a function of X = λ and a function of T = λ.0202

And hopefully you can solve these ordinary differential equations separately for capital X and T remember X that is a function of little X the variable little X T is a function of little T.0220

If you can solve both of those then if you rewrite our original guess remember was new of X and T = X of X and Multiply by T a T so, if you can find X X of X and T of T and you can put them back together and you can compile your solution to the original partial differential equation by multiplying together the solutions to these 2 ordinary differential equations.0237

So, that is the idea of separation of variables it is definitely something that will make more sense after we get some practice so, let us try that out on some examples.0271

So, in example 1 we are going to separation of variables to convert the following partial differential equation into 2 ordinary differential equations so, remember our guess for all of these for all of these separable partial differential equations is U of X T = capital X of X x T of T .0280

It looks like really need to know, some partial derivatives there so, U of X remember the T will be a constant so, that is X′ of X x T of T and U of X X will be X″ of X x T a T and it looks like really need U of T so, that would be now, holding the X is constant so, X of X x T sub' of little T.0308

To plug those into my differential equation so, plug-in plugged in to the pit the partial differential equation that we been given so, U of X X is X″ of X T of T + this is a little X now, U of T is X of X and T′ of T = 0 .0343

The idea remember is to try to solve this equation in such a way that you get all the X is on 1 side and all the Ts on the other so, I am in a move these terms over to the other side so, I get X″ of little X T of little T = - little X x X of X T′ of T .0376

Now, meant across the divide to get all my X on the left and all the T on the right so, capital X″ of X divided by up the - on the left - X X of X = π crust divide T′ of T divided by T of T .0406

Notice now, that I got all X is on 1 side all Ts on the other so, got a function of X = to a function of T and so, this must be constant that is the whole idea of separation of variables this must be constant.0429

So, then assented = to recall my constant λ and down to split that apart into 2 separate equations so, my - X″ of little X over X X of X X of X must be = to λ on 1 side and T′ of T divided by T of T = λ on the other side.0448

Each 1 of those I am not do a little bit of work to make him into a and ordinary differential equation so, X″ of X = a multiply the - over the other side - λ X x X of X and if I just pull the right-hand side over again X″ of X + λ X x X of X = 0 .0483

On the other side with the Ts and got T′ of T = λ x T of T and if I move that over T′ of T - λ x T of T = 0 .0518

So, I am not actually going to solve these differential equations the important thing is that I have done with the problem asked me which is to separate the partial differential equation into 2 ordinary differential equations 1 for in terms of X and 1 in terms of T so, let me summarize that here .0536

We got 2 ordinary differential equations 1 for X and 1 for T of T so, that is all we need to do for that problem me recap how we how we want about it so, we assume that U of XT = X of X x QT that is of an assumption we use for every single problem in this lecture.0559

That is always how you start out with separable with trying to separate partial differential equations than the X derivative the all the T constant take the X derivative second derivative is the same way holding T constant the T derivative you hold the X constant and take the derivative of T and you plug them back into the partial differential equation.0594

CNX″ of X x T + X x X x T′ = 0 down the work to get all my X on 1 side and all my Ts on the other that is the key part of separating a partial differential equation.0614

So, I did some algebra there and got only X is 1 place = to all my Ts but since I have a function of X = to a function of T it must be constant and so, I call that constant λ and then I just separated each of those equations the X = λ the Ts = to λ and each 1 of those I reorganized into a nice ordinary differential equation which you can then solve for X of X for T of T .0631

So, let us try out another example in example 2 were going you separation of variables on the following differential equation U of TT + U of XT + U of X = 0 so, let me start out the exact same guess that were going to make every single time for partial differential equations U of XT = X of X x T of T .0663

Then let me go ahead and find you what X is can be I Treating the T is constant so, X′ of X x T of T looking at the differential equation I am seeing which derivatives of that have to plug-in.0691

U of TT them in a skip a step here and skip right to the second derivative the X will come down is a constant every time and then will have the second derivative of T with respect to T now, U of XT that means you take the T derivative of the X derivative.0709

So, the X relative is a peer X′ of X x T of T the T derivative of that the X is now, constitutes X′ of X x T′ of T so, and take each 1 of those 3 derivatives and plug them into the partial differential equation and will see what happens there.0728

U TT is X of X T″ of T + U of XT is X′ of X x T′ of T + U of X is X′ of X x T of T = 0 .0749

Now, my goal here is to solve this in such a way that I get all the X is on 1 side and all the Ts on the other so, if I look at this is you got X′ here and here them in a factor that out so, X of X T″ of T + X′ of X T′ of T + T of T = 0.0774

Remove that term over to the other side so, X of X T″ of T = - X′ of X x that whole stuff in parentheses there in brackets + T of T and now, I am trying to get all my X on 1 side and all my Ts on the other so, I think I am going to cross multiply and divide .0807

On the left of me get keep double prime of T divided by that compound term T′ of T + T of T on the right I am getting at - X′ of X divided by X of X and I succeeded in getting all my X alongside only Ts on the other so, this again must be constant must be constant and so, it is got a be = to a constant that I am in a call λ.0841

Separate these 2 equations and solve each 1 separately so, my - X′ of X over X of X = λ and on the other side will have T″ of T x T′ of T + T of no divided by T′ of T + T a T is also, = to λ and each 1 of those I am going to sort out into an ordinary differential equation.0879

So, here I am going to get X′ of X move the - over the other side - λ X = - λ X of X and if I move that back I get X′ of X + λ X of X = 0 that is a nice ordinary differential equation there is only 1 variable in their which is little X .0912

On the right I see I got T″ of T = λ x T′ of T + T of T and if I move those terms over to the left I got T″ of T - λ T′ of T - λ T of T = 0 that is second-order ordinary differential equation in terms of T T of T.0949

And so, again I got 2 ordinary differential equations for X of X 1 for X of X and 1 for T of T so, I am done there let me go back over the steps in case you have any lingering questions.0985

U of XT were assuming that to be = to X of X x T of T that is the running assumption for all of these differential equations that were going to try to separate it is always separating it into a function of X x a function of T and the point of that is when you take derivatives U of X you just treating the T is a constant CX′ of X and in U of XT down here we take the T derivative of that so, we get X′ of X x T′ of T and U of TT is X of X x T″ of T.1010

The reason I took those 3 partial derivatives was because I was looking at the partial differential equation and so, I was planning ahead to see what lives can have to plug-in to the partial differential equation and so, I plugged each 1 of those in and then I am trying to solve it in such a way to get all the X over here all the Ts over there.1047

So, I see that I have not X′ of X on 2 terms here so, I factored that out and then if I move that over to the other side then I can cross multiply across divide to get all the Ts on the left and all the X on the right.1067

Which tells me that it must be = to some constant which am calling λ so, I separate that out into 2 separate equations = λ and then it is really 2 parallel tracks here 1 ends up with the differential equation in terms of X and 1 ends up with an ordinary differential equation in terms of T.1086

In example 3 were to use separation of variables to convert another differential equation into 2 ordinary differential equations so, let me remind you that our guests for every partial differential equation of forgot a use separation of variables is U of XT = X of X x T of T.1107

Now, looks like I am going to need the derivatives U X X and U TT so, U X X means were holding T constant and taking the second derivative of X should be to derivatives there x T of T and U TT means were holding X constant so, the X just comes down T″ of T and so, I am in a plugged each 1 of those into my differential equation there. 1930 It looks like I am not the plug-in you by itself as well so, U X X is X″ of X x T of T + U TT is X of X T″ of T + little T x the original U so, that is capital X of X x T of T = 0.1138

Now, I want to I I want to try and separate all the X from all the Ts and it looks like I got an X here and it X of X here thing a factor that term out of whenever I cancel got X″ of X x T of T + X of X x the quantity T″ of T + little T x T of T T T = 0.1198

Remove that term over to the other side so, I get X″ of X x T of T = - X of X x that compound term T″ of T + T x the original TFT and again I am going to cross multiply and divide so, thing the keep that - sin with the X so, get T1 - X″ of X divided by X so, it really divided both sides way X by - X of X. 2124 On the right-hand side I got T″ of T + little T x T of T divided by T of T and now, I successfully gotten all the X on 1 side all the Ts on the other so, again this thing must be constant and I am a call my constant λ that is kind of the universal and in the when you are using separation of variables.1242

And so, I got to a separate into 2 equations - X″ of X over X of X = λ on the 1 side and T″ of T + little T x TT of T over T a T = λ over here on the left and then I multiply my little my X of X out of the denominator and if you bring the - along with the with with the a X of X = - λ x X of X .1314

So, if I move that back X″ of X + λ x capital X of X = 0 on the right multiply my T out of the denominator T″ of T + little T x T of T = λ x T of T and if I move that term over the other side the other side of get T″ of T .1355

Now, I see about a T in each term here so, right this is T little T - λ just factoring their factoring out a T = 0 and so, what I discovered here is that my original partial differential equation just separated into 2 ordinary differential equations.1391

So, 2 ordinary differential equations for X of X and T of T remember were not solving these ordinary differential equations yet will do that in example little later.1416

We are just showing how you can start out with the partial differential equation make a good assumption of the beginning and hopefully separated into to ordinary differential equations so, we go back and recap that for you.1442

Start out with the original assumption U of XT = X of X x TT a T that is a running assumption we use that in every single example then where you go and that starts to vary on looking over the differential equation I see a many U of XX and U of TT .1455

So, for each 1 I hold the other variable constant and just take the second derivative with respect to the corresponding function and then I plugged each 1 of those into the partial differential equation that is what I did here plugged each 1 in.1473

And I want to separate the X for the Ts from each other and I noticed I got it X of X and in each of these terms so, I factor that out here and that enables me to pull that hold term that compound turned over to the other side and then we can cross multiply and divide and get all my X is on 1 side and all my T on the other.1490

So, since I got a function of X = to a function of T means of them both = constant animate a call that constant λ as usual and so, I separate those into 2 equations in terms of λ and 1 involves X and 1 involves Ts but there is no mixing between the 2 .1514

So, each 1 of those I can do a little algebra and rearrange it into a traditional ordinary differential equation 1 in terms of X and 1 in terms of T.1535

So, in example 4 were in you separate separation of variables on the heat equation which I said is out 1 of the most important partial differential equations and it is can be the 1 that were to spend a lot of time solving later we still have not completely solve the heat equation so, to take us a couple lectures to catch there.1545

So, were going to start solving that right now, are going to apply separation of variables to heat equation and there is a little note here that says will we get the step of using λ we should use - λ instead of λ for the separation constant that really make sense why we can do that at this stage but hopefully in a couple problems down the road from now, you will start to see why we worry is - λ instead of λ.1568

So, let us go ahead and try her separation of variables on this 1 starts out just like all the others where you start out with U of XT you assume that it can be written as X of X x T of T and then were to need some derivatives here U so, T X constant so, get X of X x T′ of T and will also, need U X X which means withhold T constant and will get the second derivative of X and then just T of T .1597

So, were to take these and plugged them in were to plug them in to our partial differential equation so, what we get here is U of T is capital X of X x T′ of T = α ² on a really care what out is at this point X″ of X x T of T and my goal here is a separate the X on 1 side the Ts on the other.1634

It is pretty easy for this 1 just get down X of X divided by X X everyone separate alone across multiplied by the other way around so, get my keys on let T′ of T divided by TT a T = anything to bring my α ² along with my T up at that down there and on the other side of got X″ of X divided by X of X .1671

By the way it is not so, obvious at this point y-axis but the α ² with 1 terms or the other there is a lot of different places you put it I am putting it here because I am kind of looking forward to a solution were going to be studying later for the heat equation.1706

So, do not worry about it right now, while putting out for ² in this term as opposed to over here with the X just bare with me and you will see will we get to problem later that it is it is a nice convention to have the α ² with the Ts.1719

Now, this is a constant because precisely because we got all X is on 1 side we got all Ts on the other and it says were going to use - λ instead of the usual λ th= - λ .1733

Remember we can untangle each 1 of those and try to I get some nice differential equations on the X side we get X″ of X divided by X of X = - λ.1750

So, if I clear my denominator there I get X″ of X = - λ x X of X and if I move it over to the other side I get X″ of X + λ x X of X = 0.1767

On the other side I am going to just keep my T′ of T divided by α ² x T of T this can only that = to - λ I think that is actually going to be the format there is going to be useful to solve later so, just can leave it in that form and what that means is that I have successfully separated my partial differential equation into 2 ordinary differential equations .1792

1 in terms of X and 1 in terms of T. I see I left out a′ on my X here; that is X″ of X + λ x X of little X = 0 and here is my differential equation in terms of T. So, it reduced my partial differential equation into two ordinary differential equations .1822

X and T now, you want to hang onto these ordinary differential equations because were going use them in the next example let us go ahead and try to solve these ordinary differential equations in the next example so, make sure you understand where these differential equations are coming for the so, the to be ready to understand them and be ready to solve them and use them in the next example.1845

So, let me go back and remind you where they came from we start with this generic assumption U of XT is a product of a function of X and a function of T and then we take its derivative with respect to T and with respect to X twice so, each time you take a partial derivative it means hold the other variable constant and you take a derivative with respect whichever variable that is.1871

So, U of T hold the X constant and you take T′ of T U of double X that means you hold the T constant and you take the second derivative with respect to X so, we plugged those back into the differential equation and then we try to sort out the X on 1 side and sort out the Ts on the other side.1896

So, we do sort those out X it turns out to be fairly easy get them sorted then we can say that the whole thing there both = to the same constant and for reasons that will be useful later it does not really make sense at this point recalling that constant - λ .1916

So, this separates out into a function of X = - λ and was we solve that into a regular ordinary differential equation and then a function of T is = λ got a little squished function of T = - λ and just do not need that in that form because that is the form that were getting used to solve later .1934

So, the next example you are going solve these 2 ordinary differential equations and were to see what kind of answer we might get to the original partial differential equation.1960

So, let us go ahead and start working on that so, were going to solve the 2 ordinary differential equations below that came from the heat equation from the partial differential equation that was the heat equation.1969

I gives us a little hint here we want to assume that λ is + to try to find solutions that satisfy the boundary conditions U of 0T = U of LT = 0 so, were to try find solutions that also, satisfy that.1985

So, let me go ahead and try to solve this first 1 this is in terms of T this is actually really nicety separable first order differential equation will my α ² over to the other side so, I got T′ of T divided by T of T = - λ α ² .2003

Now, I can integrate both sides can integrate both sides here now, the integral because I have T′ of T over T of T the integral that is exactly natural log of T of T and that is both the = to integrate the right-hand side it is were integrating with respect to T so, is just - λ α ² T + a constant .2027

I want to solve for T here so, when I raised each of both sides so, get T = E to the - λ α ² T + a constant but remember this is the same as E - λ α ² T x EC and that EC were just think of that as being a K being another constant K and so, my function T actually pretty easy to find here was a constant x E - λ α ² T.2056

Let us hang onto that were to be using it later so, before we go on to the next slide let me recap what happened on this slide to try to solve each 1 of these ordinary differential equations that come up in the heat equation and so, far just try to solve the 1 for T try to solve 1 for X on the next slide .2097

I am just trying to solve the 1 for T so, I move the α ² over to the other side key observation here is that the derivative of T of T of natural log of T would be 1 over T x its derivative so, the integral of T priority over T of T is very easy it is just natural log of T of T.2117

Integrate the right-hand side since were integrating with respect to T I just get - λ α ² x T + a constant remember it is very important to add a constant when you are integrating.2141

By the way if it is been a while since you watch the initial lectures here on educator on differential equations might be good to go back and check those out because I am using stuff that we learned back in those early lectures on differential equations .2154

So, to get rid of the natural log I raised E of both sides here and now, by laws of exponents remember E of the X + Y = E X x E Y so, since I got a + C here that is been like multiplied by E the C is just another constant so, I called a K and now, if you solve for T we get K x E - λ α ² x T.2168

So, we still have not looked at the equation for X at all yet so, that is organized go ahead and do on the next slide so, in were still working on example 5 here we solve the equation for T so, that that is looking good we have to solve the equation for X .2198

Now, this is a second order linear ordinary differential equation we had a whole lecture X a 3 lectures on how to solve these earlier on in the differential equations lecture series here on educator.com so, if you do not number how to solve those here is a really quick review.2217

You look at the characteristic equation R² + λ = 0 and using of this X of X you thing that is being X to the 0 derivative so, put R the 0s that is why there is no R right here .2235

So, were solving trying to solve for λ so, R ²s = to - λ and R = the ² root of - λ the + or - there let me remind you that we were given an assumption here which was that λ that was bigger than 0 .2254

For that means that - λ is less than 0 and so, that means the ² root of - λ is complex so, were dealing with complex numbers here it is I x the ² root of + λ.2275

So, we can have to deal with complex solutions here now, we did talk about what happens when you get complex solutions the characteristic equation that was back in 1 of earlier lectures here on educator.com back and check it out if you do not remember but the short version is that your answers look like sins and cosine.2297

So, our R here is + or - I x root λ and so, my solution X of X is I think in our earlier lectures in C1 and C2 C1 x the cosine + C2 x sin call it A and B for my constants this time so, it is a x well remember if you have a + b I then you go E AT x cosine of B T + C2 η T x sin of BT.2317

Sum up all that format simply in a change around what I am calling my constants little bit instead of C1 I am going to use a and B for C1 and C2 this a is just 0 so, you get E 0T which is just 1 so, those terms are both 1 and I get cosine of well my B is the ² root of λ so, I may go constant x quickly go ahead and call that C1 and C2 so, will be too confusing.2369

C1 x cosine of the ² root of λ and that is my variable here is X not T so, put it X in there + C2 x sin of the ² root of λ X and let me go ahead and change that C1 and C2is and call them a and B.2400

So, a cosine the ² root of λ X + Bsin the ² root of λ X so, that is my X of X and my U my my original function that I was looking for U of XT let me remind you what format that had U of XT was = to X of X this is our assumption.2424

We try to separate the variables x T of T and I have already figured out my T of T there is right there and I figured out my X of X so, let me combine those so, that = a cosine the ² of λ X + B x the sin of the ² root of λ X all x K x E - λ α ² T .2459

By the way that K drop it in the reason going to drop it is because if I put a K right here it could be absorbed right into the A and B so, just a cancel that K out and put E- λ α ² x T that is my U of XT .2491

But I am trying to satisfy my 2 boundary conditions so, let me remind you what those were my first boundary condition was U of XT are sorry U of 0T we label as a boundary condition is probably pin a while since you have seen it is a boundary condition this is given to us as part of the differential equation well as part of the package along with the differential equation was that U of 0T had to be = to 0 .2510

So, let us see what we get number that means your plug-in X = 0 let us see what we get if we plug-in X = 0 into our U of XT here so, U of 0T = a x cosine of 0+ B x sin of 0 and all of that gets multiplied by E of the - λ α ² T and that is what is to come out to be 0.2547

So, cosine a 0 is just 1 sin of 0 is 0 and so, what we get is a x E - λ α ² T = 0 on the strength of that we can say that a must be = to 0 .2576

So, we will not be needing to use that a term now, let us look at how were going to incorporate the other boundary condition and let me remind you what that was so, my U of XT simplifies down a bit when U of XT since our a term is gone away it is just B x the sin of root λ X all of that multiplied by E - λ α ² T .2596

And were told by our other boundary condition let me remind you this came as part of the package when we originally got the differential equation so, our other boundary condition was that U of L to had to be 0 remember that represented plug-in X = L which was the right-hand end of the Rod in the heat equation.2636

So, let us plug-in X = L we get B x sin of root λ L x E - λ α ² T = 0 now, we do not want it said B = 0 because otherwise our entire solution disappears .2670

Also, our E - λ α ² T we do not want that to be 0 because and even function can never be 0s that is never 0 I do not want to set B to be = to 0 because otherwise my whole solution disappears and so, what I am forced to look at is the idea that sin of root λ L = 0 .2692

So, I have to figure out what my possibilities there are to make sin of root λ L = 0 so, sin of what = 0 will I know, that sin of any multiple of π in the integer multiple of pies = to 0 so, this will work if I take root λ L = an integer multiple of π where N is an integer .2719

And so, the only thing remember L was the length of the Rod in the heat equation so, that is a constant it is established all the way through we can change that so, do not think I can solve for his is λ and so, if I will be L to the other side and ² both sides 4618 I get λ = N² π ² over L ² .2750

This will work for any value of N we choose so, we get 1 solution for each N so, I am and I write my X sub and of X remember I had an X for what I am discovering is that for each value of NI get a different solution .2784

So, and there is a B there the B could be anything so, do not know, yet what I want B to B but BN sin I have a root λ L there so, remember root λ is N π over L sin of root λ X so, N π X over L and my T = E - now, λ this is rather messy N² π ² over L ² .2813

I still have not α ² T in my exponent for E so, let me extend that α ² T such my T of T and so, ultimately get 1 solution for each end my U of an of XT is remember you multiply the X of X and the T of T so, multiply these 2 things together it is B sub N sin of N π X over L E - N² π ² α ² T all divided by L ² .2859

Pretty messy that is my solution to the heat equation that is still compatible with both of the boundary conditions that you tend to get package along with the heat equation so, that was quite a mouthful of me go back and recap everything came from.2914

We started out with well we had to solve these 2 ordinary differential equations for T of T and X of X so, we solved T of T on the previous slide previous slide so, on the previous side we figured out a form for T of T and we figured out that it was a constant x E - λ α ² T that constant I am X the going to omit that constant and the reason is because down at this step and gets multiplied by X which has Constance attach to it.2937

Anyway so, I let that K be absorbed into the Constance for the the the X functions here so, not to worry about that constant what I do have to focus on next is solving this ordinary differential equation for X and I learned how to do that back in our lessons on second-order differential equations we have an earlier lecture on that read here in the differential equations lecture series on educator.com .2980

So, if you do not remember how to solve those just check back the earlier lecture and you will see that we did a bunch of second-order differential equations and we had the characteristic equation we had cases where there were real roots where there were complex roots where there were repeated roots so, were through all those different equated different cases in different lectures.3012

What happens with this 1 is we get R ² + λ = 0 so, R ² is - λ and so, R turns into the ² root of - λ and were given that λ that was greater than 0 so, - λ is less than 0 so, really trying to here is take the ² of a - number which means we have complex solutions .3039

We learned before that will we have complex solutions A+ B I this is the format of our solution except word I using slightly different variables here so, you translate a little bit this C1 and C2 I a relabeled a and B and that is a little confusing as it is not the same as this a and B.3064

This a and B will be a was just 0 because it is 0+ or - I root λ and the B this be was root λ so, that is where I got to cosine e of that is the B there and that is the B there also, are variable now, is X instead of T we use T before but now, are using axis and so, and I also, wanted to change the Constance C1 and C2 a want to call them a and B a new a and B .3088

So, it is a little confusing with a different variables I apologize for that there is nothing very deeply mathematical going on I am just kind according the solution from the earlier lecture earlier lecture on complex roots on complex roots.3121

So, we have now, found our solution for X of X but we want to match the boundary conditions so, we want to compile our solution together with our solution from T of T so, remember this is our guess with separation of variables.3141

U is the product of X of X and T of T so, here on multiplying them together and are boundary condition was the U 0T = 0 which means you plug-in X = 0 so, I plug-in X = 0 here and that when the plug-in X = 0 you get the cosine a 0 is just 1 the sin of 0 is just 0 and so, you get a x E the something = 0.3162

What you do something this term is never = to 0 so, we must have a = 0 which means or hold term here drops out of our solution so, were just down to the simpler solution Bx the sin term x that exponential term .3191

Now, we bring in our second boundary condition which says that when you plug-in X = L then you must be getting 0 so, if we plug-in X = L here then we get complicated expression = 0 now, the E term again can not be 0 that that can not be 0.3213

I do not want B term to be 0 because it B = 0 in my entire solution disappears and I just get 0 so, I do not want to be the 0 which means I must have sin = to 0 .3214

So, if sin = 0 then I have to think about what angles have sin 0 and I remember that sin of any multiple or any integer multiple of π = 0.3248

That means root λ L must be N π for some integer and an exit is works no matter what and is as long as it is an integer sorted get a whole family of solutions with different integers if I saw for λ I get λ is N² π ² over L ² and I am just going to plug those back into my solution here back in for root λ here and back in for λ here.3261

So, the X part gives us n root λ is N π X over L the E part that the T part gives us E - λ which is - N ² π ² over L ² and we what we multiply those together and putting subscript and 0 because remember getting different solutions for each value of that so, were to have a whole bunch of solutions here .3289

So, we have BNx the X part x the T part so, that is our solution to the partial differential equation that also, satisfies both boundary conditions that is actually officially the end of this lecture here but let me go ahead and give you little Teaser for the next lecture.3318

We write down what we just figured out U of an of XT = BN x sin of N π X over L just figure this out E - N ² π ² α ² T over L ².3339

Now, let me remind you what we did there were trying to solve the heat equation and the original heat equation me make sure I write it down correctly was U of T = α ² U of X X what we have done now, as we solve that using separation of variables.3366

We also, had 2 boundary conditions U of 0T = 0 and U of LT = 0 and what were doing in this example was confirming a solution that satisfied both of those boundary conditions.3395

Now, there was 1 more condition which was the initial condition, U of X 0 = F of X and what we have not done yet is satisfied that initial condition so, that take quite a bit of work that is going to be the content of the next couple of lectures here on educator.com .3415

Were to finish off the solution of the heat equation , we are to try to make it match that initial condition so, hope you stick around and watch those lecture so, we can finish solving the equation together.3436

In the meantime, this is the end of our lecture on separation of variables, and I will just remind you that this is part of the differential equations lecture series here on educator.com. My name is Will Murray, thanks very much for joining us, bye bye.3449

Hi and welcome back to the differential equations lecture here on educator.com.0000

My name is Will Murray, and we are studying a chapter on partial differential equations.0004

We will meet the differential equations behind for this lecture and were to study Fourier series.0008

Fourier series is a tool that really used to solve the heat equation in the next lecture, but Fourier is kind of a big topic by itself so, you spent all this lecture learning about Fourier series and then the next lecture were to come back and use what we have learned about Fourier series to solve the heat equation.0015

So, you will see how that works out let us go ahead and start talking about Fourier series so, before a series for function is it is kind of like a Taylor series back at calculus 2.0034

Taylor series, your expanding things into essentially big polynomials into powers of X and Fourier series your expanding a function into sins and cosine so, near the general format here is a0 / 2 + the sum from N = 1 infinity of AN x cosine e / L + BN x sin of n π X / L so, it is it is quite complicated and will break it down will study the different elements.0045

Let me mention that this L something we seen before that is a constant and when we use this to solve the heat equation, the L is going to represent the length of the rod that were measuring the heat transfer through.0078

So, that is the length of the rod in the heat equation so, that is where we are going to get the L from now, will be a constant that stays fixed throughout the study of any particular Fourier series.0100

Everything else is something that is fixed except for these coefficients a0 and 8 AN BN so, I have to give me some formulas to find those Fourier coefficients we find them is that AN BN you find them both by these integrals it is 1 of L x the integral from - L to L of FX x now, 4 and its cosine e of n π X / L the X and for BN it sin of n π X / L DX.0111

So, were to be going lots of integrals in this lecture quite a bit of integration by parts you often end up having a new integration by parts we are finding Fourier series.0146

So, if you are rusty on integration by parts you might want to go back and review that a little bit of me doing the integration by parts kind quickly here can have to or will never get through the lecture.0158

So, you want to be up on your integration by parts we do have a lecture series here on educator.com on the second semester , calculus are on that BC calculus and there is a whole lecture in there on integration by parts of the first lecture that series.0169

So, if you do not number how to do integration by parts you might want to go back watch that lecture and when you are really up to speed on integration by parts then come back and join us for the rest of this Fourier series lectures, we will be using integration by parts quite a bit.0185

So, the Fourier coefficients we do find them using integrals sometimes these formulas simplify a little bit slower show you how they get how that can happen.0200

N = 0 there is no B 0 because the B0 would be the sin of 0 X which is just 0 so, that sort of drops out before when N = 0 the formula for a 0 simplifies because remember our formula for AN was 1 / L x the integral of - L to L of FX x cosine e of N π X / L DX.0211

That was our formula now, if N = 0 then you are just taking the cosine e 0 so, that just turns into 1 and so, the formula for a 0 just turns into the 1 / L x the integral from - L to L of FX DX .0245

That simplifies nicely another note that we have to be aware of is that all of these Fourier series are periodic so, you have to start with a periodic function and it has to be it has to have period to L.0260

That that means is that the function has to repeat itself every 2 L and usually will be considering functions that are defined from - L to L so, if you have a function that is defined maybe like that from - L to L then that is a span of 2 L so, after that it has to repeat itself so, we can only look at functions that are periodic with 2 L.0275

Usually what that boils down to is will just look at functions that are defined from - L to L and then we will just assume that there periodic outside of that interval because then we can find their Fourier series and everything will work nicely .0315

So, we have to work with periodic series for with periodic functions for Fourier series . 0331

There is a special case here when we talk about even and odd functions so, let me remind you what an even and odd function is.0339

This is a definition we probably had at summer back at calculus but you might not remember it so, how it goes is a function is even if F of - X is = 2 F of X for all X and its - X is = 2 - FX .0345

So, classic examples there are sin and cosine e sin of X for the graph of sin X their sin of X and it there is a value of X and there is - X you notice the - X is down below the x-axis where FX was above so, it is exactly been - of FX .0361

So, that is why sin of X is I cosine e of X is the other way around cosine e of X if you look at a particular value of X and then you look at - X - X you see that the cosine e of - X is the same as cosine e of X so, the cosine e satisfies this relation here F of - X is = 2 - F of X sin satisfies this relation here.0387

So, cosine e is even sin is odd X is odd and well that is not so, interesting to me maybe look at some more adjusting 1is X ² this assignment here sin X this as cosine e X look at X ² is even .0422

That is because if you look at - X this come out the same when you take their ²s X cubed on the other hand is odd so, there is X cubed and if you look at X and if you look at - X you get opposite values when you take X cubed or - X cube.0446

That is exactly why there named odd and even functions is because even powers of X X ² X forth and so, on those or even functions odd powers of X X cubed XX of the fifth and so, on those are those are odd functions that is why we use the words odd and even for those types of functions .0473

The 0 function as both even and odd because the 0 function is just a horizontal line now, there is a very nice geometric way you can identify odd and even functions and even function let us look at look at cosine e of X is her prototypical even function or X ² it those of both even.0494

You notice that if you drawn mirror on the y-axis then these functions are symmetric in this mirror their kind of symmetric about around the y-axis .0515

So, an even function has mirror symmetry across the y-axis and odd function on the other hand, look at her prototypical odd functions sin of X and odd function and X cubed is an odd function what those functions have in common is that if you put a pin in the graph right at the origin and then spun the graph around 180° you would end up looking the same as it did when you started.0527

Same thing here with X cubed if you put a pin and span around 180° it would look the same as when you start it so, we say that the graph has for an odd function has rotational symmetry around the origin that is how you can identify even and odd functions .0561

Even functions and this mirror symmetry and odd functions have this rotational symmetry lets us see how even and odd functions affect Fourier series.0580

If F is even, then it is fairly easy to check that F of X x sin of n π X / L since sin of X is odd will be an odd function which means it will had that rotational symmetry which means when you integrate from - L to L you are going to get 0.0593

So, the BN which remember we got the BN by integrating from - L to L is 1 / L there F of X time sin of N π X / L so, that is how we got the BN but if you had that rotational symmetry than the - parts in the + particular cancel each other out .0615

Which can come out to be 0 on the other hand the AN are going to have remember original formula was 1 / L from - L to L F of X cosine e of N π X / L that is going to be an even function and so, you can get the same area on the left is you are on the right.0643

Which means you can save a little time in by integrating from 0 to L and then just multiplied by 2 so, that is how that 1 turned into a 2 and that - L turned into a 0 just cutting off the - part and were doubling the graph reusing symmetry there .0668

The exact opposite thing happens if F is odd then X x cosine e of N π X is odd so, it be AN turns out to be 0 and the Fourier series contains only sins and then we take the BN we can change that original - L with 0 and we can double it .0687

So, again were exploiting that symmetry BN = 1 / L x integral from - L to L of F of X sin of n π X / L and so, because it turns out to be symmetric if F itself is odd then we can just cut it and integrate from 0 to L and then multiplied by 2 in order to fix that .0712

Then, all the AN comes out to be 0 so, if you have an odd function or even function it really cuts down on the work of finding a Fourier series have the coefficients come out to be 0 and the other half you have the sort of slightly simplified formula to solve for those coefficients.0739

So, even and odd functions are very nice for Fourier series. What is also nice about Fourier series is that if the function you are interested in is only to find from 0 to L and let me remind you that we did have that situation in the heat equation.0755

For the heat equation we we talked about this in the last couple lectures is where you are looking for the temperature of bar and you assume that the bar is oriented so, it is left denizen 0 it is right and is it L and that function FX was the initial temperature throughout the bar.0780

So, you really only had that as a function from 0 to L if you had that situation than what you can do is you can extend that function from 0 to - L so, let me squeeze me a graph in here there 0 there is L and maybe have a graph of F of X like that .0800

You really do not know, what FX does between 0 - L or maybe it is not even the fine so, you do not know, you do not have a definition for F of X from 0 to - L what you can do is you can create your own definition of F of X and you can define it in such a way that it ends up being either even or odd .0827

So, if you want to cosine e series then you would extend it to be even let me show what the picture will look like for that so, here 0 here is L and here is - L remember we have FX predefined from 0 to L but if we want it which we can do find ourselves on - L to L .0852

Define it to make it an even function from - L to L so, what I really did there was a look to X and I looked at - X and I said F of - X is defined to be the same as F of X this: = by the way what that means is defined to be.0876

So, what we did there was redefined F of - X to be the same as F of X and that forces it to be an even function so, the graph turns out to have symmetry across the y-axis now, let me show you what we could do if we wanted to get a sin series .0901

So, again there 0 there is L and suppose we have FX there and again we do not have a defined yet between - L in 0 and what I want to do is to find it to be an odd function so, does not agree, does not look quite symmetric, only do that again .0924

You see how the way had to fin d this function is if you spin it around the origin if you put a pin in the in a piece of paper at the origin event and then not spun around 180° this function would look the same in terms of equations what I did was when I looked at F of X and then for - X I defined F of - X to be - F of X .0926

Again that : there means I am defining F - X right now, to be - FX and so, in the end I get this function that is on function its graph has that rotational symmetry and what that means is that if I find a Fourier series for it really end up just using sins .0993

It turns out that when we want to solve the heat equation which is something order to talk about in the next lecture really need to use sin series so, we will extend our F of X to be odd.1013

Something we started to hit that in the previous lecture when we decided that our solutions to the partial differential equation all had sins in them and did not have any cosine so, when we try to solve this later on were going to extend functions to be odd so, that we find a Fourier series will be using only sins.1029

So, this is quite a lot to digest I think it is time with the practice some examples so, let us start with example 1 we have to find a Fourier series for the function below to start out by graphing this function.1051

So, the graph of this function looks like it is essentially defined between -2 and 2 between -2 and 0 FX = 0 graph this.1072

So, FX is = 0 between -2 and 0 and then between 0 and 2 FX is = 2 X so, there is FX is = 2 X and after that it looks like F of X +4 FX -4 is just defined to be F of X so, there is an open hole there .1086

What that means as that this function is repeating every 4 units is periodically 4 so, there it is that would be that would be 4 that would be 6 and and then - direction we make it periodic in the - direction as well.1117

And that would be - 4 -6 so, we have a function here that is periodically period 4 and want to find a Fourier series for its our L is remember As always 2L so, our L is going to be 2 because 2L was = 2 4.1144

So, the L were getting used Fourier series is 2 so, let us go ahead and find our Fourier coefficients.1173

Our first 1 is 0 we use the Fourier equations that we started with at the beginning a 0 is always 1 / L x integral from - L to L of F of X T X now, in this case the L is 2 so, this is 1/2 now, the integral from - 2 to 2.1181

You notice that from -20 the function is just 0 so, I am just a start this integral going from 0 to 2 that is because the function is 0 on the - part of that interval and FX on that interval is just X so, this is X the X now, that is a really easy integral that is 1/2 x the integral of X is X ² / 2 were integrating that from X = - 02 X = 2.1205

So, we get 1/4×2 is 4-0 so, just 1 so, I found my first or my 0 Fourier coefficient a 0 is = 1 so, I am going to hang onto that and then I am going to keep using that for my for when I want to set up my Fourier series later that if I know, a lot of other coefficients in the meantime.1241

So, I am going to use the next screen to do that so, let me remind you of everything we did on the screen first I just looked at this definition and I grasped my function it was 0 between -2 and 0 then it turned into FX = X and in this tells me it is periodically period 4.1271

Such a started repeating that function after that and I want to find my first Fourier coefficient so, I use my formula for a 0 here and that just sets up an integral. 2147 It started at it 0 but that is because this first part of the function is 0 so, would not contribute anything to the integral.1292

That I just flow through the integral and I worked out that a 0 was = 2 1 so, let me go ahead and find some more Fourier coefficients for this function .1312

Here is the function again so, now, are going to find our general AN we had at a 0 was always kind of a special case so, you want to suffer that we separately the general AN is will remember it is always 1 / L x integral from - L to L of F of X x cosine of n π X / L .1327

That was the formula we had at the beginning of the lecture and in this case our L is 2 s now, remember this function is 0 from -20 so, really do not need to worry about that part to start the integral and 0 and go take the integral from 0 2 of now, are between 0 and 2 our function is X so, FX = X x cosine of n π X / L .1354

Now, here is the first of our many bouts of integration by parts if you do not remember your integration by parts you really want to review that before you get too far down into the rabbit hole a Fourier series .1385

So, let me quickly do I use tabular integration works great for these kinds of functions so, quickly run through integration by parts here and that this is not looking familiar maybe go back and watch the lecture from calculus 2 the AP calculus BC part and you will find you will find a whole lecture on integration by parts.1400

So, right X and cosine of n π X / L and make little chart here and take derivatives on the left so, 1 and 0 integrals on the right and the integral of cosine is sin of n π X / L you get very good at doing this chart divided by N π / L .1429

So, that is the same as L / n π in order to plug in my L is = 2 I think it will make that a little easier so, my 2 plug-in L = 2 everywhere here so, 2/ n π here.1463

Down the integral cosine is sin the integral of sin is - cosine - cosine of n π X / 2 and again I have to multiply by 2 / n π so, put those together get 4 / N² π ² and now, I do little diagonal lines and alternating sins + and - and now, I can read off the answer .1481

The answer to the integration so, is my 1/2 from above and now, I am going to multiply along these diagonal lines that I got from integrating by parts so, I see I have 2 / n π X x sin of n π X / 2 so, + 4 / N ² π ² x cosine of n π X / 2 and I need to evaluate that from where to where.1515

From 0 to 2 so, from X = 0 to X = 2 so, I still have 1/2 here when I plug in X = 2 I noticed that the sin of n π X / 2 that which is the sin of n π so, that will be 0 and X = 2 so, + 4 / N ² π ² x cosine of N π x 2 / 2 so, cosine n π now, X = 0 I get 0 for the first term and for the second term I get - well cosine a 0 is 1 so, 4 / N² π ².1559

If I distribute that 1/2 then I see I got 2 / N ² π ² x cosine of n π -1 now, I have to think about what cosine of n π is and if think about it cosine of 0 is 1 cosine of π is -1 cosine of 2 π is 1 cosine 3 π is -1 in general and just alternating between 1 and -1 .1616

Cosine of n π is = 2 - while we say cosine of n π is = 1 if N is even and even and -1 if N is odd so, what we see is that if N is even then this term is going to come out to be 0 because we just get 1-1 so, 0 if N is even .1661

And if N is odd then we got a - 1 here for cosine of n π so, -1-1 is -2 so, - 4 / N ² π ² if N is odd so, quite a complicated formula here that is very typical for Fourier series will have a lot of formulas like that.1700

So, that is what I have for my AN coefficients so, let us hang onto that of this is AN and were to go on the next slide and were to figure out what BN is.1724

Let me recap what we did for AN I use my generic Fourier series formula AN is 1 / L x integral from - L to L but I can't simplify this formula because I do not have an even more and odd function for F of X here so, formula simplifies sometimes right is 2/ L x the integral from 0L and that does not work for this 1 because we do not have an even and odd function.1742

So, my L was 2 so, I plugged in 2 for all the L here the reason I put in 0 there is because the first half of this function is 0 so, and so, I should really only need to look at the interval from 0 to 2 of X that is from this X right there that is where that is coming from .1771

Cosine of n π X / 2 now, to solve that I integrated by parts I use my tabular integration trek and that expands into this rather unwieldy expression so, plug-in X = 2 X = 2 and these 2 terms came from X = 0 .1792

Fortunately sin of N π turned in the 0 that is what you get this term terms into 0 cosine of n π no such luck were just up with it the X here makes this term be 0 and then cosine of 0 is just 1 so, that is how we get that term right there .1811

Now, let us see what I do here 1/2×4 is where I got that 2 distributing 1/2 and I had N ² π ² in the denominator cosine e N π -1 cosine e N π worked out a pattern for the cosine here it is 1 if N is even an -1 if N is odd .1834

So, if it is even I just got 1-1 that is why get my 0 here if N is odd got -1-1 which is -2×2 is - 4 so, that is right at that - 4 / N ² π ² if N is odd.1861

That is all just to figure out AN still to figure out BN so, will go ahead on the next slide and will take a look at BN so, still working with the same function here we go to figure out BN so, my BN is to this formula this generic Fourier series formula - L to L of F of X x sin of n π X / L DX .1877

So, in this case our L is 2 that is 1 / 2 now, - L would be - 2 but were told that the function is 0 when were at that between -2 and 0 1 just can integrate from 0 to 2 of F of X is X x sin of n π X / L DX.1906

Fill in L is = 2 and again you can use integration by parts to integrate X x the sin of n π X / 2 so, I am using mild tabular integration trick X through the -X 1 through 0 the integral of sin is - cosine of n π X / 2 and we have to multiply that by 2 / n π by the chain rule.1941

And the integral of cosine is sin that - N π X / 2 multiply by 2 / n π again so, get 4 / N ² π ² and now, I am going to write diagonal lines and attach alternating sins that + and - .1977

So, if I get that integral all the way through what I end up with is -2 / N π x X x cosine of n π X / 2 - it looks like + 4 / N ² π ² x sin of n π X / 2 and this whole thing I need to evaluate it from X = 0 to X = 2 .2001

Plug-in X = 2 get 1/2 now, if X is = 2 I get I get -4 / N π x cosine of n π X / 2 so, that is just cosine e N π and then if I plug-in X = 2 to sin of n π X / 2 the sin of n π which is 0 and our X = 0 on the left since I got a factor of X there is just 0 .2041

On the right sin of 0 is 0 so, that turns into 0 and so, this whole thing simplifies down to -2 / N π now, cosine e N π had really like to find a good pattern for that and we did on the last slide remember cosine e N π was 1 if N is even and -1 if N is odd and there is a simple way we can write that we can is right that is -1 to the N.2082

So, I can simplify my cosine e N π is -1 to the N and what I get is -1 there is another - outsides so, -1×1 x 2 / N π so, that is my coefficient BN for my Fourier series and remind you on the previous size we already figured out the AN.2125

So, let me pull up those my a 0 was 1 we did that several slides ago AN we figured out was more complicated it was 0 if N is even and -4 / N ² π ² if N is odd .2151

So, let me assemble a Fourier series it is going to be quite complicated but let me remind you of the general formula first the Fourier series for a function is a 0 / 2 + the sum from N= 1 to infinity is a generic formula from the beginning of the lecture.2187

AN x cosine of n π X / L cosine up N π X / L + BN x sin of n π X / L so, I have figured out all the coefficients down just can drop them in this series and I think the pattern is too complicated to express it in a closed σ notation so, instead what we do is just right out the first few terms of the series.2210

In this case our Fourier series is the N = 0 term is just a 0 / 2 so, that is 1/2 now, let us look at the N = 1 terms if N is odd my AN is - 4 / N ² π ² so, from N = 1 I get -4 / π ² x cosine of n π while π X / 2 .2250

Now, my B term is -1 to the N +1 so, that is + since N = 1 from working out the N = 1 terms here my BN term is - 1 of this 2 set +2 / π 2 / N π some doing N = 1 x sin of n π X / L so, that is just π X / 2.2283

N = 2 looks like I get a break the even A is 0s finding nothing there within the odd 1 will be the BN will be - 1cube so, - 2 / 2pie will simplify that because they gets easier to see the pattern if we do not simplify that.2315

2/ 2 π x sin again I am in the right 2 π X / 2 even we know, that is would simplify 2 π X / 2 so, that is my N = 2 term .2345

Remember there was no cosine term because it it was 0 there is going to be N = 3 so, for N = 3 my AN where is -4/9 π ² - 4/9 π ² cosine of 3 π X / 2 and for my sin its +2 / 3 π sin of 3 π X / 2 .2360

So, that is my Fourier series for that function really quite complicated quite a bit of work to figure out each coefficient there so, let me remind you where I each of those elements came from we are finally told with that example.2395

So, we had already figured out a 0 in the generally ends of the previous slide just figuring out the BN so, BN was 1 / L x this integral formula that we had from the beginning of a lecture plug-in L = 2 everywhere and in that 0 was because the function was 0 on the - values there .2426

When that X comes from that X right there that we had to do some integration by parts so, here is me doing integration by parts over here using tabular integration so, it turns out to be this rather complicated formula .2450

So, we plug-in X = 2 and that is X = 2 the 2 terms we get for showing 1 of them 0 and even more fortunately we plug-in X = 0 both terms drop out and so, just get cosine e N π and I noticed from this pattern from the previous slide the cosine e N π is just - 1 of the end.2465

So, put - 1 of the end here for cosine e N π so, that is my BN there was another - here so, that combine with this -1 BN and that is why we got -1 + 1 here and did some finance -1 and +1×2 / N π.2490

So, is my BN and I just recalled from the previous slides the a 0 and the AN the a 0 and the AN and so, what I was doing now, is I am recalling the general formula for Fourier series is always has this formula a 0 / 2 + the sum of AN cosine n π X / LB & AN π X / L .2509

And so, then I fill N = 0 just gave me and a 0 / 2 give you 1/2 there and the N = 1 terms I filled in N = 1 everywhere here and I got my AN's in my BN's there and there that is how I got a 1 that is how I got B1 and then my A2 was 0 so, that is B2 . 2540

My A2 was 0 fortunately that is same a little bit there that is my a 3 right there that is a 3 and that is my B3 and at that point I ran out of space so, I decided to say that that is my Fourier series of course keeps going for infinitely many terms there .2571

So, were going use this Fourier series in the next example so, even though it is not rather complicated you do not want to completely forget it so, let me let me show you how this comes about how this gets put to use in the next example.2594

So, example 2 reuse the Fourier series that we derived in example 1 to find the value of the series 1+1/9+1/5 what these are ²s of odd numbers so, this is 1 / 3 ² 1 / 5 ² of course 1 is 1 / 1 ² and so, on .2610

So, were adding up all the ²s of odd numbers and somehow the Fourier series is mostly useful for that let me show you how it works out let me recall the Fourier series that we figured out in the previous example this is what we went through lots of work in example 1 to find so, having just watched example 1 pull up a chair and go through in and watch that it is quite long but that we get this good answer at the end of it.2634

We get the Fourier series of our function was 1/2-4 / π ² cosine of π X / 2 +2 / π this is all coming from the previous example sin a π X / 2 π X / 2 -2 / 2 π wrote it that way so, be easier to spot a pattern that is what and cancel there.2665

Sin of 2 π X / 2 -4/9 π ² cosine of 3 π X / 2 and +2 / 3 π sin of 3 π X / 2 and it kept going but 3 π X / 2.2694

The series kept going but we did not pursue it through any more terms there and the trick we want to use that for in this example is we want to plug-in X = 02 both sides there for me just remind you what this function was back from example 1 this is a function that was just a horizontal line from -2 to 0 and then was FX = X from 0 2.2720

There is 2 there is 0 there is - 2 so, that was the function were looking at so, what were delivered a plug-in X = 0 into the original function and to the Fourier series so, F of 0 that is the original function should be = 2 the Fourier series of 0 and let us see what happens there.2757

The original function F of 0 is = to 0 so, it is good 0 on the last out for a series of 0 is much more complicated so, let us see what we get when we plugged in X = 0.2782

The nice thing here is that all asine terms are going away to 0 because sin a 00 that can go to 0 that 1s can go to 0 with the balance can go to 0%.2799

All the cosine terms the cosine terms you go to 1 so, that is 1 right there that is 1 right there and so, what were going to have on the right is 1/2 -4 / π ² -4/9 π ² what is happening is were going up by ²s of odd numbers the next 1 is 4 / 25 π ² and so, on .2817

What we can do is we can manipulate this series a bit of move all the 4 / π ²s / 2 the other side and I see that I can factor out the 1 + 1/9 +1/25 and so, on and also, have that 1/2 left over on the right-hand side.2844

So, I took all of these terms and I move them over to the other side and so, trying to solve for the value of this series so, what I see is if I multiply both sides by π ² / 4 and I will get 1 + 1 / 9 +1/25 and so, on is = to π ² / 4 x 1/2.2873

And so, if I write this in σ form the way you keep track of odd numbers is you do 21 + 1 so, the sum of the reciprocals of the odd ²s is exactly π ² / 8 really quite surprising that it should come out to be such such a nice number there π ² / 8. 2900

So, we have succeeded in using that Fourier series to find the value of this curious infinite series 1 / 1 ² +1 / 3 ² +1 / 5 ² so, let me recap how we made that happen first thing I did here was I just copied the Fourier series that we derived in example 1 .2909

So, I didn't figure this out on the spot I went through lots of work back in example 1 to solve that all that Fourier series are just copied at this time and so, that what I did with that was a plug-in X = 0 I did that to the original function as a graph of the original function again that is coming from example 1.2954

Example 1 so, that is coming from example 1 so, I plug-in 0X = 0 into the original function there is right there and I also, plug it into the Fourier series while the original function we just get F 0s = 2 0 in the Fourier series we look through this see all asine terms drop out because sin of 0 is = to 0 all the cosine terms turned into 1 and so, were left with just these all these while 1/2 is kind of a special term .2977

Then all these terms with 4 / π ² and then odd-numbered ²s of odd numbers 11 925/13 ² 5 ² and so, we move those / 2 the other side we get 4 / π ² a factor that out x 1+1/9+1/5 and I still have that 1/2 over on the right and so, I am trying to solve for this odd series .3014

What I did was I multiply both sides by π ² / 4 there and so, that gave me on the right-hand side π ² / 4 ×1/2 which turned in a π ² / 8 of the left I have exactly the series that I wanted and so, I know, that my series is = 2 π ² / 8.3042

Anyone hang onto this series really use it again in the next example so, I do not forget the the value of this series so, in example 3 we have to plug-in the endpoint value X = 2 to the Fourier series above and then we want to figure out what the Fourier series converges to when the original function is discontinuous .3062

So, let me remind you what that original function looks like so, that original function it was 0 from 0 to from -2 to 2 and then we made it we made it repeat after that so, we write my scale in 2 and write my function in red .3088

So, -20 there is 2 there is 4 there is 6 there is 8 so, we made this function periodic so, there is a periodically extended function and so, the question is what happens at this end points 0 between right of the endpoint 2 what does the Fourier series do.3119

So, in order to answer this we need to go back and look at the Fourier series on the copy that down the Fourier series of X this is what we figured out in example 1 so, I am not writing this right now, this is a lot of work that we did in example 1.3158

So, 1/2 -4 / π ² cosine of π X / 2 +2 / π x sin of π X / 2 -2 / 2 / 2 π x sin of 2 π X / 2 I did not want to cancel the 2s just because I think it was easier to find a pattern that way -4/9 π ² cosine of 3 π X / 2 +2 / 3 π .3175

Cosine is sin sin cosine sin okay 2/ 3 π sin of 3 π X / 2 and then it kept going but we did not work out any farther than that in example 1 so, what I want to do is plug-in X = 2 to that Fourier series X = 2 give this X = 2.3224

Now, the Fourier series of X = 2 is = 2 1/2 now, X = 2 means that all the sins will have sin of 2 π X / 2 so, that will be sin of π sin a π is 0 sin of 2 π is still given the 0 and the sin is going to be 0 .3259

Cosine of π x / 2 so, returning to -1 here we have cosine of 3 π x 2 / 2 so, cosine 3 π is also, edition cross it off as it is really canceled but all write it as -1 here when we plug-in X = 2 remember that can be -1 .3290

So, what were going to get is 1/2 + because we have a - here - so, + 4 / π ² here we have + 4 / 9 π ² and you cannot see too much of a pattern here what is happening is getting perfect ²s of odd numbers in the denominator.3332

The next 1 is going to be 4 / 25 π ² and so, on like that so, that is what is happening there and we can simplify this event we can factor out the 4 / π ² 4 / π ² x 1+1/9 +1/25+ 1/9 next will be 49 and what she noticed here is that is exactly that odd series that we figured out what the value was back in example 2 so, you did this in example 2 .3356

Work it out on the spot here but example 2 we figured out that this was exactly = to π ² / 8 and so, what we have here is 1/2 + 4 / π ² x π ² / 8 and that simplifies really nicely 1/2 π ² cancel 4/8 is 1/2 and so, what we get for the Fourier series value it 2 is exactly 1 and let us go back and look at our original function here.3395

The original function went to let me draw a larger version of that the original function went to was FX = X so, it went up to 2 here and then because it was periodic it had to drop down to 0 again and so, what the Fourier series is doing is it says that the value is 1 which means the Fourier series is exactly picking the point exactly halfway between the left and right hand limits.3444

so, what we have maybe do that in green to really emphasize that the Fourier series so, the Fourier series order tries to mimic the function is best to can does a good job were the function is continuous but when the function is discontinuous sort of has to make a jump and what it does is it splits the difference between the left hand limits and the right hand limits .3495

So, it takes a point exactly in the middle a converges to 1 so, summarize that in words the series splits the difference and converges to appoint exactly halfway between the left and right hand limits of the original function F so, the original function F is what I got wrap there in red and the Fourier series is grasped there in green .3531

So, you can see that the green 3 much matches the red function as long as the red function is continuous for the red function makes a jump the Fourier series takes a value exactly halfway in the middle.3608

So, let us recap that example here first while I wrote out the Fourier series that we developed in example 1 I did not redo the work there and then we plug-in X = 2 which made all the sins turn out to be perfect multiples of π integer multiples of pies all the sins went to 0 all the cosine we had odd values of π cosine of π here is cosine of 3 π so, is alternately -1 so, we got 1/2 the -1 cancel out with this - .3621

So, we got + 4 / π ² + 4 / 9 π ² + 4 / 25 π ² and if you factor out 4 / π ² we got these odd on ²s building up on the bottom there we figured out in example 2 figure out what that series converges to a converges to π ² / 8 .3657

So, that is why plug-in π ² / 8 here which really nicely cancel with a 4 / π ² just end up with 1/2 +1/2 and it gave us 1 and what that is telling us is that the Fourier series were the original function the red 1 is discontinuous the Fourier series is converging to a point which is exactly halfway in between the left and the right hand limits of the original function.3681

So, example 4 were going to extend the function below in such a way that it is Fourier series will contain only cosine so, let me go ahead and draw a graph of this function.3709

FX is = 3 - X where X and between 0 and 3s with only to find between 0 and 3 sets 3 not 1 and FX = 3 - X that the line there is 3 X0 FX is 3 and is just a line straight down.3723

Try to make that line little more straight those pretty shaky lines kind of a rough angle for me as a little bit better so, there is our FX would like to extend it so, that it is Fourier series will contain only cosine .3748

Now, if you want Fourier series contain only cosine that means that you want that function to be even and so, we have to we wants F of X to be even in orders contain only cosine .3769

So, we extend and so, I am going to extend it in a way that has symmetry in the Y axis mirror symmetry and the y-axis so, that means we put my extension and read here.3792

So, that is how I am going to extend F of X to make it symmetric in the y-a Xis now, that would have to be -3 and I see that that line right there that is has slope of 1 and y-intercept of 3 so, that is why he = X +3 .3813

So, the way I am going to define my function my extended function is F of X is = 3+ X +3 4 -3 less than or = 2 X less than or = 2 0 and then I continue with the original definition 3 - X for 0 less than or = 2 X less than or = 3 and then after that I want to extend it just to make it periodic.3833

Since my L is = 3 so, I will extend it to have period.3872

What he was is 1. 2L so, in this case 2 L is 6 I will extend its at period 6 F of X -6 or X + 6 is going to be the same as F of X + or - X if X is not in between -3 and 3 so, that is my extension F of X.3883

And we go ahead and draw the periodic portion of that or that the part that extends periodically so, what that means is I have to find it from -3 to 3 and everywhere beyond that I just repeat.3913

So, that the 6 9 12 and behind that I am going to repeat the same way so, there is -6 -9 and so, on.3938

So, now, I have got a periodic function so, it will have a Fourier series and it is even it symmetric around if we just put a mirror down on the y-a Xis you can see that it symmetric on both sides of the y-axis so, it is Fourier series will contain only cosine .3971

So, we just quickly recap as soon as I see the word cosine on thinking of an even Fourier series so, I want FX to be even so, that is why I extended it to look like it has mirror symmetry in the y-Xis and that is how I got this first dotted red line here and the equation of that red line is X +3 sets wide to find my function to be X +3 from -30 .3991

And in 3 - X is the part I was already given between 0 and 3 and then I just keep on extending it whenever X is not in that interval I just add or subtract 6 from X to go back and check what he did within the interval and define it to repeat itself every 6 units after that.4021

So, hang on to this 1 because were going to use this example in the next and that the next problem and were actually going to find a Fourier series for this function see how that works out .4042

So, in example 5 were to find a Fourier series for the function that we defined back in example 4 some this comes from example 4 draw quick graph of that.4057

There is the graph of its periodic after that so, we got this dysfunction that we defined back in example 4 and we want to find a Fourier series for that contains only cosine we are the extended it to make it an even function so, we do know, that it is going to contain only cosine so, I do not even need to worry about the B ends in the Fourier formulas just getting used the AN.4070

And let me remind you of the formulas for those the Fourier coefficients the AN in general is 1 / L x the integral from - L to L of F of X x cosine of n π X / L DX but we figured out that if the function is is even then this formula simplifies a little bit you can run it from 0 to L and then due 2 / L .4101

And in the same stuff on the inside F of X x cosine of n π X / L DX and so, what I meant to do now, is figure out the coefficients A the first 1 is a 0 and we use the second form 2 / small my L is = 3 so, 2/3 x the integral from 0 to 3 of now, between 0 and 3 the function is 3 - X so, 3 - X the X.4138

I didn't worry about the cosine because cosine when N = 0 cosine e 0 is = 1 so, that is why did not worry about that part so, I get 2 thirds x now, the interval of 3 - X is 3X - X ² / 2 evaluate that from X = 02X = 3 so, 2/3 3X is when X = 3 is 9 9-9 is 9/2 so, 2/3×9/2 the 2s cancel and we just get 3 .4174

So, that is my a 0 term my other ANs are going to be a lot more complicated my other ANs are going to be 2 thirds x the integral from 0 to 3 of 3 - X x cosine of n π X / 3 DX .4224

Need a lot more space to work that out the integration by parts again so, let me take a different slide to work that 1 out let me just quickly recap what we learned on this slide.4247

My AN this is my generic Fourier series formula but we learned that if you have an even function then it simplifies a little bit you can just run it from 0 to L and by symmetry you can just double your answer so, 2 / L instead of 1 / LF of X x cosine of n π X / L when N = 0 the cosine disappears because it just turns into 1 .4262

FX is 3 - X and so, now, we get a pretty easy integral to find the 0 coefficient there just worked through the integral and it turns out to be 3 for a 0 in general AN I just took this formula I plug-in L is = 2 3 and a plug-in FX is 3 - X so, will take up that integral on the next .4287

Let us go ahead so, let me remind you what we figured out on the previous slide there we figured out that our general AN what we forgot a 0 first of all was = 3 got that on the previous side are general AN is 2/3 x the integral from 0 to 3 of 3 - X x cosine of n π X / 3 DX .4312

I want to integrate that but that is going to require integration by parts some set up my tabular integration here 3 - X and cosine of n π X / 3 so, I take derivatives of 3 - X so, that so, the derivative of 3 - X is -1 and -1 is 0 and I take integrals of cosine of n π X / 3 so, the first integral is sin of n π X / 3 .4342

And multiply that by 3 / n π by the chain rule now, the integral of sin is - cosine - cosine n π X / 3 and I multiply that by another 3 / n π so, 9 / N ² π ² like a little bit squished in their let me rewrite that.4384

9 / N ² π ² and then I do + and - some multiply down those diagonals there and so, what I get is 2/3 now, fairly complicated here 3 - X x 3 / n π x sin of n π X / 3 .4411

I see 3 - here negative so, it is the whole thing is - -9 / and ² π ² cosine of n π X / 3 and I am supposed to evaluate that whole thing from X = 0 to X = 3.4442

So, nice thing here is I see about this complicated expression right here if I plug-in X = 3 or X = 0 into this get a whole multiple of n π inside the sin .4467

So, that terms can go to 0 on both sides there for X = 3 or for X = 0 so, that is really nice on the other side it is it is not so, good so, I still got 2 thirds now, for the cosine term -9 / N² π ² x cosine of n π 3 / 3 such as cosine of n π +9 / N² π ² x cosine 0 which is just 1.4483

So, let me close that off thing to factor out the 9 / and ² π ² in 2/3 of that would give me 6 / N² π ² and then I got 1 here and - cosine of n π and let us remember the cosine of n π we worked this out earlier .4522

Cosine of n π is well when N is even then cosine e N π is 1 and when N is odd it is -1 so, if n is even then cosine of n π is just going to be 1 so, we get 1 - 1 = 0 so, this is 0 if N is even and if N is odd this then cosine e N π is - 1 we get 1 - -1 which is 3 so, 12 / N² π ² if N is odd.4553

So, that is my expression for A of N if 0s N is even and it is 12 / N ² π ² if N is odd.4604

So, let me remind you the generic form for Fourier series of X for the Fourier series of of a function it is a 0 / 2+ the sum for N= 1 new infinity of N cosine of n π X / L + BN sin of n π X / L .4620

This when we partly arrange the function so, that all the sins dropped out because we made the function be even so, I do not need to worry about the sins all those dropped out but I do still have all these cosine terms.4646

So, let me fill in what I have for the cosine terms a not that is up there and not / 2 his 3/2+ now, look to get for N = 1 that is an odd 1 so, I get 12 / π ² cosine of 1 π X / 3 so, π X / 3 N = 2 is even so, I have 0 there .4662

0 there N = 3 odd one so, 12/9 π ² and ² π ² x cosine of 3 π X / 3 go ahead and write it out that way so, it is easier to see the pattern 3 π X / 3 .4695

N = 4 is even that was 0 N = 5 is 12 / N ² π ² so, 25 π ² cosine of 5 π X / 3.4714

I see that I got a certain pattern here of odd terms so, and I try to combine those and maybe make a new series here so, I can keep track of odd terms I can start N = 0 and just used 2N + 1 to keep track of all the odd numbers that I see.4726

So, think I can factor out 12 / π ² 12 / π ² from everything there and then on the denominator all have an odd number ² so, 1 / 2 1 + 1 ² x cosine of an odd number so, 2N + 1 x π X all / 3 and that is the last word on my Fourier series using only cosine is for that function.4749

So, let me recap everything we did there we figured out that a 0 is = 3 that was on the previous slide so, not doing that again.4789

The AN we got this formula on the previous slide for my AN and of course that is in integration by parts problem so, I went to my integration by parts table here this tabular integration by parts and I worked out derivatives of 3 - X integrals of cosine e N π X / 3 and I connected them up a little diagonal lines with alternating sins .4804

Put it all together got this free horrible expression for the integral of the nice thing is that the sin term when the plug-in X = 3 or X = 0 it is can go to 0 the cosine term is not so, good if you plug-in X = 3 get cosine e N π X = 3 and if you plug-in X = 0 you get cosine e 0 which is 1 .4829

But it is being subtracted its - but it is also, the lower limits we get 2 minuses cancelled to give us a + there so, have 2/3 x looks like a factored out the 9 9×2/3 is 6 and I factored out the N ² π ² so, the 9 2/3 gave me the 6 there and the N ² π ² came out we still had a 1 here and a cosine - cosine e N π .4855

We figured this out before is it is even it is 1 if and is on its -1 and so, when it is even we just have 1 -1 so, gives a 0 if N is odd we have 1 - -1 so, without the 1 - -1 if N is odd we have 1 - -1 so, we get 2 2×6 / N² π ² gives us 12 / N² π ² so, that is our coefficient AN kind of complicated.4884

Sometimes its 0 sometimes 12 / N² π ² and I remembered my general Fourier series formula a0 / 2 + N x a cosine term BN x a sin term and we remembered that for this particular 1 we had extended the function to be even it is an even function which means that all the BN are 0 so, there is no sin terms.4925

We did that deliberately because the problem assess to find a function that had all cosine and so, my a0 was 3 that came from up here and my AN I dropped in the 12 / N² π ² there is the odd 1s right there .4950

Then we dropped in 0 for the even 1s and then I tried to rewrite this because I noticed that I had exactly the odd terms well that the 3/2 is kind of a special case.4968

But I guess you have all the odd terms here 1 3 ² 5 ² so, to find a new pattern for odd terms are used to 1 + 1 generates odd numbers so, I factored out the 12 / π ² from all of them all them have 1 / 1 1/9 1/25 so, that is 1 / 2 1 + 1 ² and enumerators there is 1 π X 3 π X 5 π X that is 1 + 1 π X / 3 .4979

So, it gives me a Fourier series for that function using only cosine which is what we are asked to do.5008

So, that wraps up this lecture on Fourier series as part of the differential equations lecture series on educator.com.5021

What we are doing here is we are trying to solve partial differential equations although we did not really get a chance to look at him in this lecture we decided a lot of background on Fourier series .5029

The whole point of that is that were going use Fourier series to solve partial differential equations so, to learn about that in the next lecture where we find a solution to the heat equation.5039

So, hope you will stick around watch the next lecture which is on using Fourier series to solve the heat equation, and that will wrap up our set of lectures on partial differential equations .5051

So again, you are watching the lectures on the differential equations here on educator.com, and my name is Will Murray, and I thank you very much for joining us today, bye bye.5062

Hi, this is Will Murray of www.educator.com and we are talking about differential equations, today we are going to talk about slope fields which are also known as direction fields.0000

Those two terms mean exactly the same thing. So if you are looking at slope fields you are looking at direction fields; if you are looking at direction fields it is exactly the same thing as a slope fields.0008

Let us see what that is all about. If the idea is that if you can manipulate a differential equation into this form right here you want to have Y′ equal to some function in terms of X and Y.0018

If the idea is that if you can manipulate into that form then you can use that form to draw what is called a slope field like I said that is the same as a direction field.0033

The way you do that is you take different points X and Y and at each point X and Y so for example you might plug in x=1 and y=1 what you can do is you can plug in those values of X and Y into the function right here the f(xy).0044

And it all spill out a slope for you. It will spill out Y′ and so what you would do is you would go over to the point 1 1 here. So there is the point 1 1 and you will draw a little mark indicating the slope of that point.0069

For example, if Y′ was 2 that would tell you at that point 1 1 you are going to draw a little mark with a slope of 2 and a slope of 2 is uphill, fairly steeply uphill.0086

I will draw a little mark like that and you would do that for each combination of X and Y that you are willing to graph so you plug in maybe another combination maybe x = 1, y = 0.0099

Supposed you get from there you get Y′ = -1 you get this by plugging them to this equation right here and then what you would do is you would go over to the 1 0 and you would draw little mark with slope -1.0112

I will draw a little mark here with slope -1 and you do that for all different points here and you find little slopes for all the points and then you gradually go through and you fill up a graph with slopes at different slopes at different points and what you get is called a slope field or a direction field.0127

We are going to have some practice with that over the course this lesson you will get the idea in the long run this is something that you want to do on a computer.0147

You want to have a computer draw slope fields or direction fields for you and then the way you used them is you kind of look at them and you can see some of the solution trajectories to the differential equations so let me talk a little more about that.0156

These solution trajectories what would happen is you will have a slope field which will show lots of different slopes like this maybe something like this so you will have all these little slopes and what you do is you kind of connect those slopes up and you will draw connect them into solution trajectories.0169

So let me draw some sample solution trajectories here and here is one right there sort of following the slopes.0192

I'm drawing curves that are parallel to the slopes wherever they passed through them so here are some solution trajectories these differential equation and if we have the actual function for this curves these functions would be solutions to the differential equation.0200

What we are finding is different solutions to the differential equations one key point here is the solutions never crosses each other so you want to be careful when you are drawing the solution trajectories to understand that they should not be crossing each other.0220

They sort of locally they are always parallel to each other they might diverge or converge but they never touch each other they never cross each other.0233

Another thing that is key to understand is that these different solution trajectories correspond to different values of the arbitrary constant and the general solution to the differential equations.0242

If you remember when we solve some of these differential equations we get things like y=(ce)^(x^2) something like that might be our general solution to the differential equation and that C was an arbitrary constant.0252

That is an arbitrary constant we do not know at least initially what the value of that constant is and so if you plugged in different value of that constant you would get different graphs and those different graphs are these different solution trajectories.0268

Each one of these red curves here is a solution to the differential equation they just correspond to different values of the arbitrary constant in the general solutions.0284

To figure out which solution trajectory you want you use an initial condition which is usually given to you as part of the package with the differential equation so you will have an initial value for X0 and Y0.0297

What that does is that gives you an initial point X0 and Y0 somewhere on the graph and that will tell you which of the solution trajectories is actually the solution that satisfies not just the differential equation but also your initial condition.0315

If we had an X0 Y0 if you have an initial condition that lets you narrow it down from the whole family of curves to just a single solution trajectory.0334

That is kind of the geometric picture there algebraically we have only seen these process where we get a general solution to a differential equation it has an arbitrary constant in it and then what we do is we take this point and we plug it in to the general solution.0344

You plug in x=x0 y=y0 into the general solution to find the value of C so that is how the initial condition can be useful algebraically is you plug it in to the general solution and solve for the value of the arbitrary constant C.0366

Geometrically what you are doing is using that single initial point to differentiate between all these different solution trajectories and pick the single one that fits your initial condition.0389

I think that is enough theory I want to jump into some examples and let us see how we can draw these slope fields or direction fields in actual practice.0402

On our first example we have to draw a direction field remember that is the same as a slope fields for the following differential equation Y′=x+2y so what I'm going to do is make a chart of combinations of X's and Y's.0412

And for each one of those I'm going to plug them in to the differential equation and I'm going to figure out a slope corresponding to it and then I'm going to try to plot all those on a set of axis.0428

I'm going to make a fairly big graph here we use a pretty big scales so I'm going to make maybe right that 1, that is -1, that is -1, that is 1, so 1, -1, 1, -1, 2, 2, -2, -2 and the idea is that I'm going to plug different combinations of X's and Y's into the differential equations here.0441

And figure out what the slopes should be at those different points so I will start out easy I will start with the X and Y or 0 and 0 and if I plug in 0 0 Y′ remember Y′ is always x+2y that is from the differential equation.0474

What that means in terms of the graph is I'm going to the point 0 0 and I'm going to draw a little hash mark with slope 0 so there is my hash mark with slope 0.0489

Let me go back and calculate a few more points here if I have x=1 y=0 then x+2y it will just be 1 then if I have x=0 y=1 then Y′ x+2y will be 2 and if they are both 1 then x+2y would be 1+2x1 so 3.0503

Let me plot those so 1 0 that is on the x axis I'm going to draw a little line with slope 1 and draw mine here a little bigger the line with slope 1, 0 1 is up here, slope 2 is steeper.0527

I'm going to make that a bit steeper probably still is not steeper enough let me try a little bit steeper there we go and at 1 1 we got the point 3 so we got the value of 3 that is really quite steep I will draw that almost vertical.0543

There is a point there that is almost vertical there we go ahead and fill in some more points here that I want to find some slopes here I'm going to try to calculate some slopes at each of those points there I think.0561

I will calculate those as fast as I can see I got -1 0 and Y′ is x+2y that is a -1 there, -1 1, x+2y will be +1 there see I have got -2 0, x+2y is -2, -2 1 is x+2y that would be 0.0577

And -2 -1 I will just go across on the southern axis there -2 -1 would be x+2y would be -2-2 -4, -1 -1 would be -3 0 -1 x+2y would be -2 1 -1 x+2y -1 and 2 -1 x+2y is 0 again.0606

And see this couple of more points there 2 0 x+2y would be 2 and 2 1 x+2y would be 4 so let me go ahead and fill in all my slopes. So let's see I have done up to here -1 0.0646

I see I have a slope of -1 so let me give this a slope of -1 there -1 1 gives me a slope of 1.0665

There is the point -1 1 and I will draw a line with slope 1, that is going uphill -2 0 gives me a slope of -2 so that one is fairly steeply downhill, -2 1 gives me a slope of 0 so that is a horizontal line there -2 -1 gives me slope of -4 very steeply downhill, -1 -1 gives me slope of -3 still pretty steeply downhill.0679

0 -1 gives me a slope of -2, 1 -1 gives me a slope of -1 so it is not so steep anymore and 2 -1 gives me a slope of 0 so that is horizontal, 2 0 gives me a slope 2 pretty steeply uphill now and 2 1 gives me a slope of 4 very steeply uphill.0711

The idea is that you go through and kind of as many points is you are willing to fill in you try to find the slope by plugging them all into this differential equation and then you draw these little hash marks everywhere through there.0735

I'm going to stop there because it gets a little tedious after this but in the long run this is something that is done very effectively on a computer.0750

We are just practicing a couple by hands so that you understand the process you understand what the computer is doing when they generate these pictures.0759

What I'm going to do on the next example is show you a computer generated picture of the slope fields you will see it looks like my slope field but it got little hot more points field in.0765

So let me just remind you what we did here we took different values of X's and Y's and then we plug each one into x+2y and we found these slopes for each combination of X and Y.0777

And then we found those on the actual graph and for each one of those points I made a little hash mark with that slope for example -1 1 that means X is -1 and Y is 1 so that is that point right there is -1 1.0791

And when you plug -1 1 into the differential equation that means X is -1y is 1 so Y′ for that particular point comes out to be -1+2 which is 1 so that is why we go with slope 1 and that is how I got this hash mark with slope 1 right there.0809

What I just did was I did that for every single point here I found the Y′ and I drew a little hash mark at each place so in the next example we are going to come back and look at the same differential equation.0833

And we are going to use the slope fields we are going to draw some a solution trajectories but instead of using my crude hand drawn graph here we are going to use one produce on a computer.0847

Let us how that turns out so in example too we are going to use a slope field for Y′ +x=x+2y from example 1 that was the slope field we generated back in the previous example If you have not worked through that one.0857

You really want to check that one out so that you understand these examples in the next slide what I'm going to do is give you a computer drawn slope fields so it will be a little bit easier for me to work with it.0872

What we will do with it is we will draw some solution trajectories to the differential equation we are going to check that this equation right here is the general solution to the differential equation.0884

I will actually plug it into the differential equation and we will check and make sure that it works that it is the general solution and then we look back at our solution trajectories.0896

And try to understand which one corresponds to which values of that arbitrary constant so that is our arbitrary constant right there.0904

We will try to figure out which solution trajectories corresponds to which values of that arbitrary constant in the general solution.0912

Let us take a look at the computer drawn slope for this differential equation here it is and this is something we started to draw by hand but I got tired after plotting I do not know 15 or so points.0919

That is why I pull out the computer drawn example the differential equation we are using was Y′=x+2y and what we have to do here is look at these direction fields.0935

These direction field remember it comes from example 1 that is how you figure out where this picture came from is if you have not just watch example and go back and check out example 1.0949

You will see that we learn how we generate this direction field, what I'm going to do is draw some solution trajectories here which means I'm going to kind of go through and connect the dots.0960

Except that the curves I draw have to be parallel to these little slope marks everywhere I draw them so I'm going to draw these solution trajectories in ready.0970

I see that there is kind of a dividing line here If I just follow those straight arrows right there I get one solution trajectory.0981

That is the easiest one to draw and I'm going to draw a few more by kind of connecting these things up in curves that are basically whenever I touch an arrow I should be parallel to that arrow.0992

Let me draw a few more of these whenever I go near an arrow I need to be following the slope dictated by that arrow.1005

Here they start more gently but they still and up going uphill and all the ones above this line here seem to be ending up going uphill.1012

Once below the line I have to kind of turn myself to make this work if I follow this arrow I got to be parallel to the arrow every time I touch an arrow I have to be parallel on it.1026

I'm sort of forced to be constantly downhill if I start below the line here so those are my solution trajectories the point of this is that anyone of these red curves would be a solution to these differential equation.1040

Anyone of these red curves because they are obeying the slopes at every single point that means that they are obeying the differential equation at every single point so there are solutions to the differential equation.1059

Every one of these red curves is a solution to the differential equation now we are asked to given a general solution to the differential equation let me write that again here.1072

That was Y is equal to -1½ x I'm just copying a slope to previous line -1½ x-1/4 plus arbitrary constant e^2x we are asked to confirm that that is a solution to these differential equation.1083

What I'm going to do is I'm going to plug that in to the differential equation and I'm just going to see if it works so I need to calculate Y′ the derivative of -1 ½ x is -1½ -¼ is a constant so it is derivative is just 0 so +0 and then ce^2x the derivative of that is 2ce^2x.1100

I see I can simplify that, that is -1½ +2c e^2x and so that is Y′ let me work out the right hand side of the differential equation.1124

The right hand side is x+2y let me label that as RHS for right hand side so that is the right hand side of the differential equation I'm going to see if I get the same thing as I got from Y′.1137

RHS is x+2y now that is x+ and I'm going to plug in my Y from up here so 2x-1½ x- ¼ +ce^2x and let me just simply that X now 2x-1½ x is -x, 2x-1 ¼ is -1 ½ and then I guess I got + 2ce^-2x.1153

And those X is cancelled let me just simplifies down to -1½ +2ce^2x and if you check back up here that Y′ and the x+2y when we work them out those are equal to each other.1187

We have a solution to the differential equation we have a solution to the differential equation we found the general solution to the differential equation and I got that just by plugging it in to the differential equation and checking.1210

The last thing we are asked to do is identify which of the solution trajectories correspond to which values of C in the differential equation so let me write the general solution to the differential equation one more time y=-1½ x - ¼ +ce^2x.1233

Now let me plug in some different values of c here if I plug in c=0 then I just get y is equal to -1½ x - ¼ that is a line that is just it is in y = mx + b form that is a line.1255

And it is a line with -1½ slope and I see that right here in the middle of my graph this line right here in the middle so let me flag this solution trajectory this line is the line corresponding to c=0.1273

That is the solution trajectory corresponding to c= 0 now what if I have a C value that is not 0 if I have C value that is positive then what I get is Y is equal to well that same equation for the line +ce^2x where that ce^2x is positive.1298

What I'm doing is I'm taking outline and I'm adding an exponential function to it which means remember e^2x let me do a little graph of e^2x, e^2x looks like this so I'm taking outline and I'm adding an exponential function to it.1323

That is going to give me this positive exponentially curving upwards solution trajectories so all of these solution trajectories correspond to positive values of the arbitrary constant in the general solution to the differential equation.1339

Now if c < 0 I will get Y is equal to a line +ce^2x but this time the C would be negative so essentially if I do -e^2x the graph of -e^2x will be curving sharply downhill.1361

I'm taking that line and I'm adding on a term that sharply downhill and that is how you get these negatively curve solution trajectories downhill below that line.1385

All these solution trajectories down here correspond to negative values of the constant in the general solution to the differential equation.1397

I have done everything I was asked to do here we are supposed to take the direction field remember we figure out the direction field in example 1 so if this graph that I started out with all the little hash mark that is a total mystery to you.1410

Go back and look at example 1 you will where it comes from except that in example 1 we just did it by hand and here I produced the nicer computer generated one.1424

Looking at this direction field I was able to connect up the hash marks and draw a solution trajectories the whole idea of that was that wherever I drew a curve it had to be parallel to these arrows at wherever was near an arrow.1435

It could not cut through an arrow like that or like that it is got to be parallel to these arrows wherever it sees an arrow so no cutting the arrows it is going to fall the arrows.1449

I started out by noticing that somebodies arrows give me a straight in the middle and then if I sort of connect the dots above the line I got the solutions trajectories occurring upwards.1461

Below the line I get all the solution trajectories that supposed to be more parallel to the arrow that would have to curve down.1471

The next thing I had to do there was to check the given general solution and make sure that that really was the solution to the differential equation.1484

In order to do that I worked out Y′ on one side and I worked out x+2y independently so here is me working out Y′ took the derivative and then I worked out x+2y where for the Y I plugged in what are solution is so I plugged that in for a solution.1494

Simplified it down and lowing behold It did check I did get the same thing as we got up here for Y′ so that's nice that those two things did check each other out that shows that we do have a solution to the differential equation.1512

But finally the problem asked us to determine which values of the arbitrary constant corresponded to which solution trajectories and I started easy by taking the constant equals 0 which just simplified me down there something in y = mx + b form so that was just Y equals a line.1530

This line right here it is c =0 this line right into the middle and then I checked if C is not 0 then we are doing is we are either adding or subtracting an exponential function to our line.1549

Here is my typical graph of an exponential function goes upper down depending on what you are adding or subtracting and so we are taking that line and so these more complicated solution to a diverge from that line.1562

By either going uphill if c > 0 or by going downhill if c < 0 so I figured out which values which kind of values of a constant will correspond to which family of solution trajectories.1574

Let us keep moving and try another one in example 3 we are going to draw to a direction field for the following differential equation Y′ is equal to -x over y.1591

I'm going to start by making a little chart here in which I'm going to list value combination of X's and Y's and then I'm going to plug them in to the differential equation and I'm going to see what I get for Y′.1603

Once I got a good number of point stored up I'm going to draw a graph of those and see if I can get a general idea of what the direction field would look like.1623

There is 1, there is -1, -1 that one is a little far out let me draw that a little closer there is -1 and there is 1, 2, 2 and so on.1633

What I'm going to do is start with different combinations of X's and Y's and try to plug them in to the differential equation.1655

I will try 0 and 0 coz that is all is good one to start with unfortunately if I try to plug that into -x over y I get 0 over 0 which is undefined so I really do not get anything useful there.1662

I will try x is 1 and y is 0 and then I get one over 0 actually -1 over 0 which I think of that it is being infinity so I think of that is being infinite slope, I think of that is being a vertical line.1678

At 1 0 I'm going to draw a vertical line here let me try 1 1 and 0 1 if I plug in 1 1 I get -x over y -1 over 1 is equal to -1 so the .1 1.1694

Draw a few more points here then I'm planning to fill in the .1 1, I just figured out has slope of -1 so I will draw a line with negative slope there.1716

The .0 1 if I plug that in to -x over y I get -0 over 1 which is 0 let me draw some more points I here -1 0 gives me 1 over 0 which is I think of that is being infinity so I think of that is another vertical line.1736

-1 1 will give me 1 over 1 plugging that in that is going to give me 1, -1 -1 will give me +1 again because it is 3 negative signs there wait sorry -1 -1 let me write this out so many negative signs It is hard to keep track -1 over -1.1758

Three negative signs will give me -1, -1 0 -1 will give me 0 over -1 it will give me a 0 again 1 -1 will give me a +1, 2 2 gives me -2 over 2 is in -1.1785

Actually I did not want to do 2 2, I did not want to go that far hang on I will just going to do 2 1 which gives me -2 over 1 which is -2 over 1 is -2, -2 1 gives me 2 over 1 which is +2 and -2 -1 gives me 2 over 1 which is, I'm sorry -2 over 1 coz there are two negatives and then a third negative is -2 and 2 -1 will give me 2 over 1 which is 2 again.1813

Let me go ahead and try